The acid strength is characterized by their acid-ionization constants. The larger the K_a, the stronger the acid. Notice that it is opposite for the pK_a – the smaller the pK_a, the stronger the acid.

Among these, HCl is the strongest acid with a K_a of 1.3 x 10⁶. The second strongest would the sulfurous acid, K_a (H₂SO₃) = 5.9 x 10^-2, and the weakest acid here is the protonated ethyl amine, K_a (CH₃CH₂NH₃⁺) = 2.5 x 10^-11.

The acid strength is characterized by their acid-ionization constants. The larger the K_a, the stronger the acid, and therefore, for a weaker acid, we are looking for a smaller K_a.

HClO₃ is a strong acid so it completely dissociates in an aqueous solution.

K_a (HCN) = 6.2 x 10^-10

K_a (HC₂H₃O₂) = 1.8 x 10^-5

Based on the K_a values, HCN is the weakest acid.

The pH is defined as the negative logarithm of proton/hydronium ion concentration:

(a) pH = -log 1.3 x 10^-2 = 1.9

(b) pH = -log 1.6 x 10^-3 = 2.8

(c) We can determine the pOH first, and then, using the pH and pOH correlation, calculate the pH.

pOH = -log 1.8 x 10^-3 = 2.7

pH + pOH = 14, therefore, pH = 14 – pOH

pH = 14 – 2.7 – 11.3

Alternatively, we can determine the concentration of H⁺ ions from the relationship [OH^–] · [H⁺] = K_w = 10^-14.

\[\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{1.8\, \times {{10}^{ – 3}}}}\, = \,5.6\, \times \,{10^{ – 12}}\]

pH = -log [H⁺] = -log 5.6 x 10^-12 = 11.3

The [OH^–] and [H⁺] can be calculated from one other using their relationship [OH^–] · [H⁺] = K_w = 10^-14.

\[{\rm{(a)}}\,\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{2.5\, \times \,{{10}^{ – 6}}}}\, = \,4.0\, \times \,{10^{ – 9}}\]

\[{\rm{(b)}}\,\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{8.3\, \times \,{{10}^{ – 4}}}}\, = \,1.2\, \times \,{10^{ – 11}}\]

\[{\rm{(c)}}\,\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{4.6}}\, = \,2.2\, \times \,{10^{ – 15}}\]

\[{\rm{(d)}}\,\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{3.9\, \times \,{{10}^{ – 2}}}}\, = \,2.6\, \times \,{10^{ – 13}}\]

The [H₃O⁺] can be determined by the antilog of the pH. The [OH^–], in turn, is calculated from the [H₃O⁺]  and [OH^–] relationship.

a) pH = 7.20

Step 1. Determine the [H₃O⁺]

[H₃O⁺] = 10^-pH = 10^-7.20 = 6.31 x 10^-8

Step 2. Determine the [OH^–] using their relationship [OH^–] · [H⁺] = K_w = 10^-14.

\[\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{6.31\, \times \,{{10}^{ – 8}}}}\, = \,1.58\, \times \,{10^{ – 7}}\]

The solution is basic because [H₃O⁺] < [OH^–] the pH > 7.

b) pH = 15.3

Step 1. Determine the [H₃O⁺]

[H₃O⁺] = 10^-pH = 10^-15.3 = 5.01 x 10^-16

Step 2. Determine the [OH^–] using their relationship [OH^–] · [H⁺] = K_w = 10^-14_.

\[\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{5.01\, \times \,{{10}^{ – 16}}}}\, = \,20.\]

The solution is basic because [H₃O⁺] < [OH^–] the pH > 7.

c) pH = 4.60

Step 1. Determine the [H₃O⁺]

[H₃O⁺] = 10^-pH = 10^-4.60 = 2.51 x 10^-5

Step 2. Determine the [OH^–] using their relationship [OH^–] · [H⁺] = K_w = 10^-14.

\[\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{2.51\, \times \,{{10}^{ – 5}}}}\, = \,3.98\, \times \,{10^{ – 10}}\]

The solution is acidic because [H₃O⁺] > [OH^–] the pH < 7.

Strong acids and bases completely dissociate in aqueous solutions, and therefore, the H⁺ concentration is going to be the same as the concentration of the monoprotic acid.

(a) 0.15 M HCl

pH = -log [H⁺] = -log 0.15 = 0.82

(b) 0.60 M HClO₄

pH = -log [H⁺] = -log 0.60 = 0.22

(c) 1.4 M KOH

Strong acids and bases completely dissociate in aqueous solutions, and therefore, the [OH^–] is going to be the same as the [KOH].

We can determine the pOH first, and then, using the pH and pOH correlation, calculate the pH.

pOH = -log 1.4 = -0.15

pH + pOH = 14, therefore, pH = 14 – pOH

pH = 14 – (-0.15) = 14.15

(a) Strong acids and bases completely dissociate in aqueous solutions, and therefore, the H⁺ concentration is going to be the same as the concentration of the monoprotic acid.

pH = -log [H⁺] = -log 0.0025 = 2.6

(b) 0.84 M NaOH

Strong acids and bases completely dissociate in aqueous solutions, and therefore, the [OH^–] is going to be the same as the [NaOH].

We can determine the pOH first, and then, using the pH and pOH correlation, calculate the pH.

pOH = -log 0.84 = 0.076

pH + pOH = 14, therefore, pH = 14 – pOH

pH = 14 – 0.076 = 13.924

We need to determine the concentration of H⁺ ions which is equal to the concentration of HCl because it is a strong acid completely dissociated in water.

So, the first step is to calculate the moles of HCl and determine the molarity.

\[{\rm{n}}\,{\rm{(HCl)}}\,{\rm{ = }}\,\frac{{\rm{m}}}{{\rm{M}}}\,{\rm{ = }}\,\frac{{{\rm{24}}{\rm{.0}}\,{\rm{g}}}}{{{\rm{36}}{\rm{.5}}\,{\rm{g/mol}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.658}}\,{\rm{mol}}\]

\[{\rm{M}}\,{\rm{(HCl)}}\,{\rm{ = }}\,\frac{{\rm{n}}}{{\rm{M}}}\,{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.658}}\,{\rm{mol}}}}{{{\rm{0}}{\rm{.662 L}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.994}}\,{\rm{mol/L}}\]

pH = – log [H⁺] = – log 0.994 = 0.00261 = 2.61 x 10^-3

We need to determine the concentration of OH^– ions which is equal to the concentration of KOH because it is a strong base and completely dissociates in water.

So, we need to calculate the moles of KOH from the pH. For this, let’s first determine the [H⁺] from the pH and then, the [OH^–] using the relationship [OH^–] · [H⁺] = K_w = 10^-14.

Step 1. Determine the concentration of H⁺ ions:

pH = – log [H⁺], [H⁺] = 10^-pH

[H⁺] = 10^-9.80 = 1.58 x 10^-10

Step 2. Determine the [OH^–]

\[\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{1.58\, \times \,{{10}^{ – 10}}}}\, = \,6.33\, \times \,{10^{ – 5}}\]

This is the concentration of OH^– ions which is equal to the concentration of KOH.

At this point, we know the concentration and the volume of the KOH solution, and to find the mass of KOH, we first calculate its moles from the molarity:

Step 3. Determine the moles of KOH:

\[{\rm{n }}\left( {{\rm{KOH}}} \right){\rm{  =  MV  =  6}}{\rm{.33}}\; \times \,{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}\,\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\,{\rm{0}}{\rm{.6800}}\,\cancel{{\rm{L}}}{\rm{ = }}\,4.30\, \times \,{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}\,{\rm{mol}}\]

Step 4. Convert the moles to mass:

m (KOH) = 4.30 x 10^-5 mol x 56.1 g/mol = 2.41 x 10^-3 g

The first step is to determine the new concentration of HNO₃ using the dilution formula. So, let’s label the M and V for the stock solution and the new solution:

M₁ = 3.00 M 

V₁ = 40.0 mL

V₂ = 40.0 + 210.0 = 250. mL

M₂ = ?

M₁V₁ = M₂V₂

\[{{\rm{M}}_{\rm{2}}}\,{\rm{ = }}\,\frac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{M}}_{\rm{2}}}\,{\rm{ = }}\,\frac{{{\rm{3}}{\rm{.00}}\,M\, \times \,{\rm{40}}{\rm{.0}}\,{\rm{mL}}}}{{{\rm{250}}{\rm{.}}\,{\rm{mL}}}}\; = \,0.48\;M\]

This is the concentration of HNO₃, and because it is a strong acid, the [H⁺] is also 0.480 M.

Therefore, the pH is:

pH = – log [H⁺] = – log 0.480 = 0.319

First, let’s write the dissociation equation of HCN:

HCN ⇆ H⁺ + CN^–

Initially, the concentration of HCN is 0.45 M, and we can assign x M as the concentration of HCN that dissociate (change in the ICE table). The equilibrium concentration of HCN is going to be decreased by x mol/l, so it is (0.45 -x) M.

The initial concentration of H⁺ and CN^– ions are 0, but at equilibrium, they are going to be x mol/l because that is how much the HCN dissociates.

So, let’s set up an ICE table with these values:

HCN ⇆ H⁺ + CN^–

Next, we set up the equation based on the Ka:

\[{K_{\rm{a}}}\; = \,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][C}}{{\rm{N}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HCN]}}}}\; = \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.45}}\;{\rm{ – }}\;{\rm{x}}}}\; = \;{\rm{4}}{\rm{.9 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Now, we make an approximation that 0.45 – x ≈ 045 because the dissociation constant is very small and by doing so, we get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.45}}\;}}\; = \;{\rm{4}}{\rm{.9 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Therefore,

x = 1.4849 x 10^-5

Make sure the approximation is valid, and for this, x must be less than 5% of HCN initial concentration (less than 5% of the acid dissociates). 1.4849 x 10^-5/0.45 = 0.0033 %, so the approximation is valid.

So, x = 4.1 x 10^-5 , and this is the [H⁺] at equilibrium, therefore,

pH = -log 1.4849 x 10^-5 = 4.8

The answer is reasonable because we have an acidic solution and the pH < 7.

First, let’s write the dissociation equation of acetic acid:

CH₃CO₂H ⇆ H⁺ + CH₃CO₂^–

Initially, the concentration of CH₃CO₂H is 0.74 M, and we can assign x M as the concentration of CH₃CO₂H that dissociates (change in the ICE table). The equilibrium concentration of CH₃CO₂H is going to be decreased by x mol/l, so it is (0.74 -x) M.

The initial concentration of H⁺ and CH₃CO₂^– ions are 0, but at equilibrium, they are going to be x mol/l because that is how much the CH₃CO₂H dissociates.

So, let’s set up an ICE table with these values:

CH₃CO₂H ⇆ H⁺ + CH₃CO₂^–

Next, we set up the equation based on the Ka:

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.74}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.7  \times  1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

Now, we make an approximation that 0.74 – x ≈ 0.74 to get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.74}}\;}}\; = \;{\rm{1}}{\rm{.7 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

Therefore,

x = 3.546 x 10^-3

Make sure the approximation is valid, and for this, x must be less than 5% of CH₃CO₂H initial concentration (less than 5% of the acid dissociates). 3.546 x 10^-3/0.74 = 0.48 % – the approximation is valid.

So, x = 3.546 x 10^-3, and this is the [H⁺] at equilibrium, therefore,

pH = -log 3.546 x 10^-3= 2.45 = 2.5

The answer is reasonable because we have an acidic solution and the pH < 7.

The first part here is going to be determining the initial concentration of benzoic acid by converting the mass to moles and dividing it by the volume of the solution.

\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H)}}\,{\rm{ = }}\,{\rm{0}}{\rm{.86}}\,\cancel{{\rm{g}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{122}}{\rm{.1}}\,\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.0}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 3}}}}\,{\rm{mol}}\]

Because the volume of the solution is 1.0 L, the molarity is equal to the moles of the acid.

Next, we write the dissociation equation of benzoic acid:

C₆H₅CO₂H ⇆ H⁺ + C₆H₅CO₂^–

Initially, the concentration of C₆H₅CO₂H is 7.0 x 10^-3 M, and we can assign x M as the concentration of CH₃CO₂H that dissociates (change in the ICE table). The equilibrium concentration of C₆H₅CO₂H is going to be decreased by x mol/l, so it is (7.0 x 10^-3 – x) M.

The initial concentration of H⁺ and C₆H₅CO₂^– ions are 0, but at equilibrium, they are going to be x mol/l because that is how much the C₆H₅CO₂H dissociates.

So, let’s set up an ICE table with these values:

C₆H₅CO₂H ⇆ H⁺ + C₆H₅CO₂^–

Next, we set up the equation based on the Ka:

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{7}}{\rm{.0 x 1}}{{\rm{0}}^{{\rm{ – 3}}}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

Notice that the initial concentration is very low, and the approximation is likely going to be invalid, but to make sure, let’s go ahead and test it.

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{7}}{\rm{.0 x 1}}{{\rm{0}}^{{\rm{ – 3}}}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

Therefore,

x = 6.69 x 10^-4

The percent ionization is 6.69 x 10^-4/7.0 x 10^-3 = 9.56 %, so the approximation is not valid.

Therefore, we must solve the quadratic equation from the K_a expression.

\[\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{7}}{\rm{.0 x 1}}{{\rm{0}}^{{\rm{ – 3}}}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

x² = -6.4 x 10^-5x + 4.48 x 10^-7

Rearrange to a form of standard quadratic equation:

x² + 6.4 x 10^-5x – 4.48 x 10^-7 = 0

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = 1, b = 6.4 x 10^-5, c = -4.48 x 10^-7

Solving this, we find that:

x = 6.3 x 10^-4

So, x = 6.3 x 10^-4, and this is the [H⁺] and [C₆H₅CO₂^–] at equilibrium. The concentration of the acid at equilibrium is the difference between the initial concentration and the change:

[C₆H₅CO₂H] = 7.0 x 10^-3 – 6.3 x 10^-4 = 6.37 x 10^-3 M

 The pH is:

pH = -log 6.3 x 10^-4= 3.2

Now, let’s see how different the pH would have been if we kept the approximation. The concentration found by the approximation was 6.69 x 10^-4, so the pH is:

pH = -log 6.69 x 10^-4= 3.17

The difference is insignificant and if we round off the numbers, we get the same answer. However, it is not every time that the answers will be so close, and it’s a good idea to follow the accepted guidelines for the approximation which is the ionization should be less than 5% of the initial concentration of the acid.

To summarize, the concentrations at equilibrium and the pH are going to be:

[C₆H₅CO₂H] = 6.37 x 10^-3 M

[C₆H₅CO₂^–] = 6.3 x 10^-4 M

[H⁺] = 6.3 x 10^-4 M

pH = 3.2

First, let’s write the dissociation equation of acetic acid:

HF ⇆ H⁺ + F^–

Initially, the concentration of HF is 2.0 M, and we can assign x M as the concentration of HF that dissociate (change in the ICE table). The equilibrium concentration of HF is going to be decreased by x mol/l, so it is (2.0 – x) M.

The initial concentration of H⁺ and F^– ions are 0, but at equilibrium, they are going to be x mol/l because that is how much the HF dissociates.

So, let’s set up an ICE table with these values:

HF ⇆ H⁺ + F^–

Next, we set up the equation based on the Ka:

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{2}}{\rm{.0}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.6  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Now, we make an approximation that 2.0 – x ≈ 2.0 because, the dissociation constant is very small and by doing so, we get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{2.0\;}}\; = \;{\rm{6}}{\rm{.6 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Therefore,

x = 0.0363

Make sure the approximation is valid, and for this, x must be less than 5% of HF initial concentration (less than 5% of the acid dissociates). 0.0363/2.0 = 1.82 % – the approximation is valid.

So, x = 0.0363, and this is the [H⁺]at equilibrium, therefore,

pH = -log 0.0363= 1.4

The answer is reasonable because we have an acidic solution and the pH < 7.

First, let’s write the dissociation equation of HA:

HA ⇆ H⁺ + A^–

The K_a expression is:

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][}}{{\rm{A}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HA]}}}}\;\]

Now, if we determine the [H⁺] from the pH, we will also know the [A^–], and how much of the acid had ionized because the dissociation reaction is a 1:1 ratio.

pH = – log [H⁺], [H⁺] = 10^-pH

[H⁺] = 10^-2.68 = 0.0021

This is the concentration of the acid that is ionized, therefore, the equilibrium concentration of the acid is:

[HA]_eq = [HA]_{initial –} [HA]_ionized = 0.246 – 0.0021 = 0.2439

And now that we have all the equilibrium concentrations, we can plug them in the K_a expression and calculate its value:

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{\left( {0.0021} \right)\left( {0.0021} \right)}}{{{\rm{0}}{\rm{.2439}}}}\; = \;1.81\, \times \,{10^{ – 5\;}}\]

Like in the previous example, we can determine the [H⁺] from the pH, which will also indicate how much of the acid had ionized because the dissociation reaction occurs in a 1:1 ratio.

pH = – log [H⁺], [H⁺] = 10^-pH

[H⁺] = 10^-3.1 = 7.94 x 10^-4

This is the concentration of the acid that ionized, therefore, if we assign x M for the initial concentration of the acid, the equilibrium concentration will be:

[HNO₂]_eq = [HNO₂]_{initial –} [HNO₂]_ionized = x – 7.94 x 10^-4

Now we can set up an ICE table and solve for x:

HNO₂ ⇆ H⁺ + NO₂^–

The K_a expression will be the basis of the equation to determine the x.

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\;\]

\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{\left( {7.94\, \times \,{{10}^{ – 4}}} \right)\left( {7.94\, \times \,{{10}^{ – 4}}} \right)}}{{{\rm{x}}\; – \,7.94\, \times \,{{10}^{ – 4}}}}\; = \;7.2\, \times \,{10^{ – 4\;}}\]

6.305 x 10^-7 = 7.2 x 10^-4x – 5.717 x 10^-7

 x = 1.67 x 10^-3

And this is the initial concentration of the acid.

H₂S is a diprotic acid, and because the acid ionization constant for the second step is very small, we can ignore the H⁺ contribution from the second step.

H₂S(aq) ⇆ H⁺(aq) + HS^–(aq) K_a1 = 1.0 x 10^-7

HS^–(aq) ⇆ H⁺(aq) + S^2-(aq) K_a2 = 1.0 x 10^-19 (ignore)

Initially, the concentration of H₂S is 0.40 M, and we can assign x M as the concentration of H₂S that dissociate (change in the ICE table). The equilibrium concentration of H₂S is going to be decreased by x mol/l, so it is (0.40 – x) M.

The initial concentration of H⁺ and HS^– ions is 0, but at equilibrium, they are going be x mol/l because that is how much the HS^– dissociates.

So, let’s set up an ICE table with these values:

H₂S(aq) ⇆ H⁺(aq) + HS^–(aq)

Next, we set up the equation based on the Ka:

\[{K_{\rm{a}}}_1\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][S}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[}}{{\rm{H}}_{\rm{2}}}{\rm{S]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.40}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.0  \times  1}}{{\rm{0}}^{{\rm{ – 7}}}}\]

Make an approximation that 0.40 – x ≈ 0.40 to get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{0.40\;}}\; = \;{\rm{1}}{\rm{.0 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 7}}}}\]

x = 2.0 x 10^-4

Make sure the approximation is valid, and for this, x must be less than 5% of H₂S initial concentration (less than 5% of the acid dissociates). 2.0 x 10^-4/0.40 = 0.05 %, so the approximation is valid.

So, x = 2.0 x 10^-4, and this is the [H⁺] at equilibrium, therefore,

pH = -log 2.0 x 10^-4= 3.7

The answer is reasonable because we have an acidic solution and the pH < 7.

The first step is to write the reaction of the weak base with water:

NH₃(aq) + H₂O(l) ⇆ NH₄⁺(aq) + OH^–(aq)

Next, assume x mol/l NH₃ has reacted and set up an ICE table. Initially, the concentration of NH₃ is 0.85 M, and its equilibrium concentration is going to be decreased by x mol/l, so it is (0.85 – x) M.

The initial concentration of NH₄⁺ and OH^– ions are 0, and at equilibrium, they are going to be x mol/l because that is how much of the ammonia reacts with water.

The expression for the equilibrium constant will then be:

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}\; = \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.85}}\;{\rm{ – }}\;{\rm{x}}}}\; = \;{\rm{1}}{\rm{.8 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

This is a quadratic equation and remember a shorted way of solving it to use an approximation that 0.85 – x ≈ 0.85 because, the ionization constant of the weak is very small and therefore, the percentage of it reacted with water is negligible compared to its initial concentration.

Doing so, we get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.85}}\;}}\; = \;{\rm{1}}{\rm{.8 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 5}}}}\]

Therefore, x = 0.0039

Make sure the approximation is valid, and for this, x must be less than 5% of NH₃ initial concentration:

\[\frac{{{\rm{0}}{\rm{.0039}}}}{{{\rm{0}}{\rm{.85}}}}\; \times \;100\%  = \;{\rm{0}}{\rm{.46\%  }}\]

Therefore, the approximation was valid and the concentration of OH^– ion is 0.0039 M and we can calculate the pOH:

pOH = -log [-OH] = -log 0.0039 = 2.4

And finally, we can calculate the pH using the pH and pOH correlation:

pH =14 – pOH = 14 – 2.4 = 11.6

Alternatively, we can determine the concentration of H⁺ ions from the relationship [OH^–] · [H⁺] = K_w = 10^-14.

\[\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{0.0039}}\, = \,2.564\, \times \,{10^{ – 12}}\]

pH = -log [H⁺] = -log 2.564 x 10-12 = 11.6

A pH of 11.6 indicates a basic solution which additionally suggests that our calculations were correct.

The first step is to write the reaction of the weak base water:

(C₂H₅)₃N(aq) + H₂O(l) ⇆ (C₂H₅)₃NH⁺(aq) + OH^–(aq)

Next, assume x mol/l (C₂H₅)₃N has reacted and set up an ICE table. Initially, the concentration of (C₂H₅)₃N is 0.25 M, and its equilibrium concentration is going to be decreased by x mol/l, so it is (0.25 – x) M.

The initial concentration of (C₂H₅)₃NH⁺ and OH^– ions are 0, and at equilibrium, they are going to be x mol/l because that is how much of the ammonia reacts with water.

The expression for the equilibrium constant will then be:

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[}}{{\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}} \right)}_{\rm{3}}}{\rm{N}}{{\rm{H}}^{\rm{ + }}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[}}{{\left( {{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}} \right)}_{\rm{3}}}{\rm{NH]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.25}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.0  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Make an approximation that 0.25 – x ≈ 0.25 to get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.25}}\;}}\; = \;{\rm{4}}{\rm{.0 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Therefore, x = 0.010

Check if the approximation is valid, and for this, x must be less than 5% of (C₂H₅)₃N initial concentration:

\[\frac{{{\rm{0}}{\rm{.010}}}}{{{\rm{0}}{\rm{.25}}}}\; \times \;100\%  = \;{\rm{4\%  }}\]

Therefore, the approximation was valid and the concentration of OH^– ion is 0.010 M.

The [H⁺] can be calculated from K_w:

\[\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{0.010}}\, = \,1.0\, \times \,{10^{ – 12}}\]

Next, we can calculate the pH:

pH = -log [H⁺] = -log 1.0 x 10^-12 = 12

To summarize, we determined that:

[OH^–] = 0.010 M

[H⁺] = 1.0 x 10^-12 M

pH = 12

The formula of caffeine is C₈H₁₀N₄O₂, so we can write its reaction with water as:

C₈H₁₀N₄O₂(aq) + H₂O(l) ⇆ C₈H₁₀N₄O₂H⁺(aq) + OH^–(aq)

Next, assume x mol/l of caffeine has reacted, and set up an ICE table. Initially, the concentration of caffeine is 0.40 M, and its equilibrium concentration is going to be decreased by x mol/l, so it is (0.40 – x) M.

The initial concentration of the conjugate acid C₈H₁₀N₄O₂H⁺ and OH^– ions is 0, and at equilibrium they are going be x mol/l because that is how much of the caffeine reacts with water.

One additional step here is to convert the pK_b to K_b:

pK_b = -log K_b, K_b = 10^-pKb

K_b = 10^-10.4 = 3.98 x 10^-11

The expression for the equilibrium constant will then be:

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[}}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{10}}}}{{\rm{N}}_{\rm{4}}}{{\rm{O}}_{\rm{2}}}{{\rm{H}}^{\rm{ + }}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{10}}}}{{\rm{N}}_{\rm{4}}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.40}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.98  \times  1}}{{\rm{0}}^{{\rm{ – 11}}}}\]

Make an approximation that 0.40 – x ≈ 0.40 to get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.40}}\;}}\; = \;{\rm{3}}{\rm{.98 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 11}}}}\]

Therefore, x = 4.0 x 10^-6

Make sure the approximation is valid, and for this, x must be less than 5% of caffeines’ initial concentration:

\[\frac{{4.0 x 1{0^{ – 6}}}}{{{\rm{0}}{\rm{.40}}}}\; \times \;100\%  = \;{\rm{9}}{\rm{.9}}\, \times \;{10^{ – 4}}{\rm{\%  }}\]

Therefore, the approximation was valid and the concentration of OH^– ions is 4.0 x 10^-6 M . The [H⁺] can be calculated from K_w:

\[\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{4.0\, \times \,{{10}^{ – 6}}}}\, = \,2.5\, \times \,{10^{ – 9}}\]

pH = -log [H⁺] = -log 2.5 x 10^-9 = 8.6

To summarize, we determined that:

[OH^–] = 4.0 x 10^-6 M

[H⁺] = 2.5 x 10^-9 M

pH = 8.6

In this problem, once we set up the ICE table, make the equation, and solve for the x, we will only need to divide it by the initial concentration of the amine to find the percent ionization. This is what we have been doing when checking the validity of the approximation.

So, first, write the reaction of the weak base water:

CH₃CH₂NH₂(aq) + H₂O(l) ⇆ CH₃CH₂NH₃⁺(aq) + OH^–(aq)

Next, assume x mol/l CH₃CH₂NH₂ has reacted and set up an ICE table. Initially, the concentration of CH₃CH₂NH₂ is 0.64 M, and its equilibrium concentration is going to be decreased by x mol/l, so it is (0.64 – x) M.

The initial concentration of CH₃CH₂NH₃⁺ and OH^– ions are 0, and at equilibrium, they are going to be x mol/l because that is how much of the amine reacts with water.

The expression for the equilibrium constant will then be:

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}^{\rm{ + }}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.64}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.6  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Make an approximation that 0.64 – x ≈ 0.64 to get a simplified equation:

\[\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.64}}\;}}\; = \;{\rm{5}}{\rm{.6 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Therefore, x = 0.019

Make sure the approximation is valid, and for this, x must be less than 5% of CH₃CH₂NH₂ initial concentration:

\[\frac{{{\rm{0}}{\rm{.019}}}}{{{\rm{0}}{\rm{.64}}}}\; \times \;100\%  = \;{\rm{3\%  }}\]

Therefore, the approximation was valid, and we correctly determined that 3% of the amine is ionized.

First, let’s write the ionization equation of morphine (C₁₇H₁₉NO₃):

C₁₇H₁₉NO₃(aq) + H₂O(l) ⇆ C₁₇H₁₉NO₃H⁺(aq) + OH^–(aq)

The K_b expression is:

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[}}{{\rm{C}}_{{\rm{17}}}}{{\rm{H}}_{{\rm{19}}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}{{\rm{H}}^{\rm{ + }}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{{\rm{17}}}}{{\rm{H}}_{{\rm{19}}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}{\rm{]}}}}\]

Now, if we determine the [OH^–] from the pH, we will also know the [C₁₇H₁₉NO₃H⁺] because they are formed in a 1:1 ratio when the amine abstracts a proton from water.

To do this, we first calculate the [H⁺] from the pH.

pH = – log [H⁺], [H⁺] = 10^-pH

[H⁺] = 10^-10.9 = 1.26 x 10^-11

\[\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{1.26\, \times \,{{10}^{ – 11}}}}\, = \,7.94\, \times \,{10^{ – 4}}\]

This also represents the concentration of the reacted morphine, therefore, the equilibrium concentration of morphine is:

[morphine]_eq = [morphine]_{initial –} [morphine]_ionized = 0.340 – 7.94 x 10^-4 = 0.339

And now that we have all the equilibrium concentrations, we can plug them in the K_b expression and calculate its value:

\[{K_{\rm{b}}}\;{\rm{ = }}\,\frac{{\left( {7.94\, \times \,{{10}^{ – 4}}} \right)\left( {7.94\, \times \,{{10}^{ – 4}}} \right)}}{{0.339}}\; = \;1.86\, \times \,{10^{ – 6\;}}\]

A good shortcut to figuring out if the salt is acidic or basic, is to identify the “strong component” in it.

(a) KBr is a salt of a strong base and strong acid, and therefore, it is neutral.

(b) FeCl₂ is a salt of strong acid HCl, and a weak base Fe(OH)₂, and therefore, it is acidic.

(c) Na₂CO₃ is a salt of a strong base NaOH, and a weak acid H₂CO₃, and therefore, it is basic.

(d) Al(NO₃)₃ is a salt of strong acid HNO₃, and a weak base Al(OH)₃, and therefore, it is acidic.

(e) KClO₄ is a salt of a strong base and strong acid, and therefore, it is neutral.

(f) Na₂SO₃ is a salt of a strong base NaOH, and a weak acid H₂SO₃, and therefore, it is basic.

(g) NH₄ClO₃ is a salt of strong acid HClO₃, and a weak base NH₃, and therefore, it is acidic.

NaBr is a salt of a strong base and strong acid, and therefore, it is pH-neutral.

Although Mg(OH)₂ is not as strong of a base as NaOH or KOH, it is still considered a strong base. Therefore, MgCl₂ and Mg(NO₃)₂ are neutral because they are prepared from a strong base and a strong acid.

NH₄Cl is a salt of strong acid HCl, and a weak base NH₃, and therefore, it forms an acidic solution.

KF is a salt of a strong base KOH, and a weak acid HF, and therefore, it forms a basic, or the most basic among the given, solution.

NaOBr is a salt of a strong base NaOH and a weak acid HBrO and therefore, the solution will be basic.

To determine the pH of a salt, we dissociate it and write the reaction of its weak component with water. If the reaction produces H⁺/H₃O⁺, the solution is acidic, and if OH- ions are formed, the solution will be basic.

NaOBr → Na⁺ + BrO^–

BrO^– + H₂O(l) → HBrO(aq)  +  OH^–(aq)  (basic)

Next, determine the K_b from the K_a:

K_a · K_b = K_w = 10^-14

Rearranging this, we get an expression for K_b:

\[{K_{{\rm{b}}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{a}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{2}}{\rm{.90}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 9}}}}}}\; = \;{\rm{3}}{\rm{.45 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 6}}}}\]

Set up an ICE table for K_b assigning x mol/l for ionization.

Write the expression for K_b using the equilibrium concentration sin the ICE table in order to determine [OH^–].

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[HBrO][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[Br}}{{\rm{O}}^ – }{\rm{]}}}}\; = \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.74}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.74}}}} = \;{\rm{3}}{\rm{.4}}\; \times \,{\rm{1}}{{\rm{0}}^{{\rm{ – 6}}}}\]

Solving for x, we find that x = 0.0016 mol/l and we check if the approximation was valid:

\[\% \, = \;\frac{{0.0016}}{{0.74}}\; \times \;100\% \; = \;0.22\% \]

The approximation is valid, so the concentration of OH^– ions at equilibrium is 0.0016 M.

We can now calculate the pOH, and then pH using the pH + pH = 14 relationship:

pOH = -log 0.0016 = 2.8, and therefore, the pH is:

pH = 14 – pOH = 14 – 28 = 11.2

And this answer is also reasonable because we predicted to have a basic solution.

Ammonium chlorate (NH₄ClO₃) is the salt of a weak base ammonia (NH₃) and a strong acid HClO₃, so it yields an acidic solution.

NH₃ + HClO₃ → NH₄ClO₃

So, let’s follow what we have discussed earlier and see why it produces an acidic solution.

Dissociate the salt to identify the weak component ion that will react with water:

NH₄ClO₃ → NH₄⁺ + ClO₃^–

NH₄⁺ is the ion with a weaker component (NH₃), so it is going to react with water producing hydronium ion which makes the solution acidic:

NH₄⁺(aq)  + H₂O(l) → NH₃(aq)  +  H₃O⁺(aq)  (acidic)

The next part is to determine the concentration of H₃O⁺ ions in order to calculate the pH.

We are given the K_b, which measures the base strength and it is based on the reaction of ammonia itself with water:

NH₃(aq) + H₂O(aq) ⇆ NH₄⁺(aq) + OH^–(aq)

This equation does not contain H⁺/H₃O⁺ ions, and we do not know the concentration of ammonia, therefore, we need to determine the K_a.

K_a and K_b can be calculated from one another using the relationship:

K_a · K_b = K_w = 10^-14

Rearranging this, we get an expression for K_a:

\[{K_{\rm{a}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{b}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{1}}{\rm{.8}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}}}\; = \;{\rm{5}}{\rm{.7 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Once we know the K_a, it is the same procedure as we do when calculating the pH of a weak acid.

We assign x mol/l for ionization and set up an ICE table to determine the [H₃O⁺]:

NH₄⁺(aq)  + H₂O(l) → NH₃(aq)  +  H₃O⁺(aq)  (acidic)

Next, write the expression for K_a using the equilibrium concentrations in the ICE table:

\[{K_{{\rm{a}}}}\; = \,\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{][}}{{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}\; = \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.26}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.26}}}} = \;{\rm{5}}{\rm{.7 x 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Therefore, x = 1.2 x 10^-5

Check if the approximation was valid:

\[\% \, = \;\frac{{1.2\, \times \;{{10}^{ – 5}}}}{{0.26}}\; \times \;100\% \; = \;0.0047\% \]

The percentage of hydrolysis is less than 5%, and therefore, the approximation is valid.

The concentration of H₃O⁺ ions is 1.2 x 10^-5 M, so the pH is:

pH = -log 1.2 x 10^-5 = 4.9

This indicates an acidic solution which is what we expected to have for the solution of NH₄ClO₃.

KNO₂ is a salt of a strong base KOH and a weak acid HNO₂ and therefore, the solution will be basic.

Dissociate the salt and write the reaction of its weak component with water.

KNO₂ → K⁺ + NO₂^–

NO₂^–(aq) + H₂O(l) → HNO₂(aq) + OH^–(aq)  (basic)

Next, determine the K_b from the K_a:

K_a · K_b = K_w = 10^-14

Rearranging this, we get an expression for K_a:

\[{K_{{\rm{b}}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{a}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{4}}{\rm{.0}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 4}}}}}}\; = \;{\rm{2}}{\rm{.5 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 11}}}}\]

Set up an ICE table for K_b assigning x mol/l for the ionization.

Write the expression for K_b using the equilibrium concentrations in the ICE table in order to determine the [OH^–].

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.35}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.35}}}}{\rm{ = }}\;{\rm{2}}{\rm{.5}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 11}}}}\]

Solving for x, we find that x = 3.0 x 10^-6 mol/L, and we check if the approximation was valid:

\[\% \, = \;\frac{{3.0\, \times \;{{10}^{ – 6}}}}{{0.35}}\; \times \;100\% \; = \;0.00084\% \]

The approximation is valid, so the concentration of OH^– ions at equilibrium is 3.0 x 10^-6 M.

We can now calculate the pOH, and then pH using the pH + pH = 14 relationship:

pOH = -log 3.0 x 10^-6 = 5.5, and therefore, the pH is:

pH = 14 – pOH = 14 – 5.5 = 8.5

And this answer is also reasonable because we predicted a basic solution.

CH₃CO₂Na is a salt of a strong base NaOH and a weak acid CH₃CO₂H therefore, the solution will be basic.

Dissociate the salt and write the reaction of its weak component with water.

CH₃CO₂Na → Na⁺ + CH₃CO₂^–

CH₃CO₂^–(aq) + H₂O(l) → CH₃CO₂H(aq) + OH^–(aq)  (basic)

Next, determine the K_b from the K_a:

K_a · K_b = K_w = 10^-14

Rearranging this, we get an expression for K_a:

\[{K_{{\rm{b}}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{a}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{1}}{\rm{.7}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}}}\; = \;{\rm{5}}{\rm{.9 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Set up an ICE table for K_b assigning x mol/l for the ionization.

CH₃CO₂^–(aq) + H₂O(l) → CH₃CO₂H(aq) + OH^–(aq)

Write the expression for K_b using the equilibrium concentrations in the ICE table in order to determine [OH^–].

\[{K_{{\rm{b}}}}\; = \,\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1}}{\rm{.0}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1}}{\rm{.0}}}}{\rm{ = }}\;{\rm{5}}{\rm{.9}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

We find that x = 2.4 x 10^-5 mol/l.

Check if the approximation was valid:

\[\% \, = \;\frac{{2.4\, \times \;{{10}^{ – 5}}}}{{1.0}}\; \times \;100\% \; = \;0.0024\% \]

The approximation is valid, so the concentration of OH^– ions at equilibrium is 3.0 x 10^-6 M.

We can now calculate the pOH, and then pH using the pH + pH = 14 relationship:

pOH = -log 2.4 x 10^-5 = 4.6, and therefore, the pH is:

pH = 14 – pOH = 14 – 4.6 = 9.4

And this answer is also reasonable because we predicted to have a basic solution.