## General Chemistry

#### Practice

1.

Which of the following solutions can be classified as buffer solutions?

(a) NaBr + HBr, (b) NaHSO4 + H2SO4, (c) HCl + HOCl (d) Na2HPO4 + NaH2PO4, (e) CH3CH2NH2 + CH3CH2NH3+ (f) NaNO2 + HNO2, (g) KCN + HCN, (h) Na2SO4 + NaHSO4, (i) NH3 + NH4ClO3, (j) CH3CO2H + NaOH.

(d) Na2HPO4 + NaH2PO4, (e) CH3CH2NH2 + CH3CH2NH3+ (f) NaNO2 + HNO2, (g) KCN + HCN, (h) Na2SO4 + NaHSO4, (i) NH3 and NH4ClO3

Solution

(a) HBr is a strong acid, and therefore, this is not a buffer. Additionally, NaBr is a salt of a strong acid (HBr) and strong bases (NaOH)

(b) H2SO4 is a strong acid, and therefore, this is not a buffer.

(c) HCl is a strong acid, and therefore, this is not a buffer.

(d) Na2HPO4 is a weak base (HPO42- is the conjugate base of a weak acid H2PO4) and NaH2PO4 is a weak acid, therefore, this is a buffer solution.

(e) CH3CH2NH2 is a weak base and CH3CH2NH3+ is its conjugate acid, therefore, this is a buffer solution.

(f) HNO2 is a weak acid and NO2is its conjugate base, therefore, this is a buffer solution.

(g) HCN is a weak acid and CNis its conjugate base, therefore, this is a buffer solution.

(h) HSO4 is a weak acid, and its conjugate base, SO42 is a weak base, therefore, this is a buffer solution.

(i) NH3 is a weak base and NH4ClO3 (NH4+) is its conjugate acid, therefore, this is a buffer solution.

(j) Although CH3CO2H is a weak acid, NaOH is a strong base, and this cannot work as a buffer. In fact, they will react as soon as mixed.

2.

Addition of which salt will suppress the ionization of HOCl?

1. Na2SO4
2. KNO3
3. NaOCl
4. NaNO2

NaOCl

Solution

NaOCl will make HOCl ionize less than it does in pure water because of the common ion effect. HClO is a weak acid and establishes an equilibrium forming H+ and ClO ions:

HOCl(aq) ⇆ H+(aq) + ClO(aq)

The hypochlorite ion is also formed when NaOCl dissociates and, being a common, ion it pushes the equilibrium of acid dissociation to the left.

3.

A solution of ammonia and ammonium chloride is a common buffer system. Write the corresponding reactions to show how it resists a pH change when an acid or a base is added to it.

NH3(aq) + HNO3(aq) → NH4HNO3(aq)

NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH3(aq) + H2O(l)

or

NH4+(aq) + OH(aq) → NH3(aq) + H2O(l)

Solution

Ammonia (NH3) is a weak base, and it neutralizes any acid added to the buffer:

NH3(aq) + HNO3(aq) → NH4HNO3(aq)

NH4Cl is the conjugate acid and neutralizes any base added to the system:

NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH3(aq) + H2O(l)

or

NH4+(aq) + OH(aq) → NH3(aq) + H2O(l)

4.

Calculate the pH of the buffer solution consisting of 0.75 M NH3 and 0.95 M NH4Cl. Kb (NH3) = 1.8 x 10-5.

pH = 9.15

Solution

Approach 1 – using the Henderson–Hasselbalch Equation

The Henderson–Hasselbalch Equation can be shown as:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{A}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HA]}}}}$

Or labeling the species as acid and a base:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[Base]}}}}{{{\rm{[Acid]}}}}$

In this example, the base is NH3, and the acid is NH4+.

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}$

The initial concentration of ammonium ion is going to be equal to the concentration of NH4Cl because it is a strong electrolyte and completely dissociates into ions:

0.95                       0.95
NH4+Cl (aq) → NH4+(aq) + Cl(aq)

So, we have the concentration of ammonia and the ammonium ion, so the only missing part is the pKa.

Remember, Ka is the dissociation constant of the acid or the conjugate acid, while Kb is the dissociation constant of the base. The corresponding dissociation equations are:

NH3(aq) + H2O(aq) ⇆ NH4+(aq) + OH(aq) Kb (NH3) = 1.8 x 10-5

NH4+(aq) + H2O(aq) ⇆ NH3(aq) + H3O+(aq) Ka – to be determined

The pKa can be calculated by either first determining the Ka from Kb since Ka · Kb = Kw,  or by first determining the pKb and converting using the relationship: pKa + pKb = 14

Let’s go with the second option this time:

pKb = -log Kb = -log 1.8 x 10-5 = 4.75

Therefore, the pKa is:

pKa = 14 – pKb = 14 – 4.75 = 9.25

Once the pKa is known, plug the values in the Henderson–Hasselbalch Equation:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}$

${\rm{pH}}\; = \;9.25\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.75}}\,M}}{{{\rm{0}}{\rm{.95}}\,M}}\, = \,\,9.15$

Always check if your answer is reasonable by referring to the pH and pKa relationship:

We have pH < pKa, and this is correct because [NH4+] > [NH3].

Approach 2 – using the Equilibrium Method

For this approach, we will need to set up an ICE table and determine the concentrations to eventually calculate the pH.

We are given the Kb, so let’s write the reaction of ammonia with water:

NH3(aq) + H2O(aq) ⇆ NH4+(aq) + OH(aq) Kb (NH3) = 1.8 x 10-5

${K_{\rm{b}}}\; = \;\frac{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}\; = \;1.8\, \times \,{10^{ – 5}}$

The initial concentration of NH3 is 0.75 M, and we can set its ionization as x M. The equilibrium concentration of NH3 is then (0.75 -x) M.

The initial concentration of ammonium ion is going to be equal to the concentration of NH4Cl because it is a strong electrolyte and completely dissociates into ions:

0.95                       0.95
NH4+Cl (aq) → NH4+(aq) + Cl(aq)

At equilibrium, the total concentration of NH4+ is going to be the sum of the concentrations from both reactions:

[NH4+] = (0.95 + x) M

${K_{\rm{b}}}\; = \;\frac{{{\rm{[0}}{\rm{.95}}\;{\rm{ + }}\,{\rm{x][x]}}}}{{{\rm{[0}}{\rm{.75}}\,{\rm{ – }}\,{\rm{x]}}}}\; = \;1.8\, \times \,{10^{ – 5}}$

And now, we make the approximation that x << 0.75 and 0.95, therefore,

${K_{\rm{b}}}\; = \;\frac{{{\rm{0}}{\rm{.95x}}}}{{{\rm{0}}{\rm{.75}}\,}}\; = \;1.8\, \times \,{10^{ – 5}}$

x = 1.42 x 10-5

Check if the approximation is valid:

1.42 x 10-5/0.75 = 1.9 x 10-3% – valid

x = 1.42 x 10-5 and this is the [OH], from which we can calculate the pOH and then the pH.

pOH = -log [OH] = -log 1.42 x 10-5 = 4.847

pH = 14 – pOH = 14 – 4.847 = 9.152

So, what we see is that the pH is almost identical (it would be identical if we round off to correct significant figures) regardless of how we calculate it. Therefore, from now on, we will use the Henderson–Hasselbalch Equation to calculate the pH of a buffer. It should work for most cases when the concentrations are larger than 0.1 M, and the dissociation constants are very small. One thing to keep in mind is that if the Henderson–Hasselbalch Equation does not work, then the buffer is not suitable in the first place.

5.

Calculate the pH of the buffer solution which is 1.4 M CH3COONa and 1.8 M CH3COOH. Ka (CH3CO2H) = 1.7 x 10-5

4.65

Solution

The Henderson–Hasselbalch Equation can be shown as:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{A}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HA]}}}}$

In this example, HA is the acetic acid, and Ais its conjugate base:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH]}}}}$

Don’t get confused with CH3COONa and CH3COO – they both effectively represent the same species as CH3COONa is a strong electrolyte and completely dissociates into ions in aqueous solutions.

The only missing part in the equation is the pKa which we calculate by the negative log of Ka:

pKa = -log Ka = -log 1.7 x 10-5 = 4.76

Therefore, the pH is:

${\rm{pH}}\; = \;4.76\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{1}}{\rm{.4}}}}{{{\rm{1}}{\rm{.8}}}}\; = \,4.65$

The pH < pKa, and this makes sense because [HA] > [A]:

6.

Calculate the pH of the buffer solution that is 0.70 M NaCN and 0.55 M HCN. Ka (HCN) = 4.9 x 10-10

4.79

Solution

The Henderson–Hasselbalch Equation for this example is:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{N}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HCN]}}}}$

Don’t get confused with NaCN and CN – they both effectively represent the same species as NaCN is a strong electrolyte and completely dissociates into ions in aqueous solutions.

The only missing part in the equation is the pKa which we calculate by the negative log of Ka:

pKa = -log Ka = -log 4.9 x 10-10 = 4.69

Therefore, the pH is:

${\rm{pH}}\; = \;4.69\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.70}}}}{{{\rm{0}}{\rm{.55}}}}\; = \,4.79$

The pH > pKa, and this makes sense because [A] > [HA]:

7.

Calculate the pH of the buffer solution that is 0.60 M propionic acid (CH3CH2CO2H) and 0.75 M sodium propionate. Ka (CH3CH2CO2H) = 1.32 x 10-5

4.88

Solution

The Henderson–Hasselbalch Equation for this example is:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{CO}}{{\rm{O}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{COOH]}}}}$

Don’t get confused with CH3CH2COONa and CH3CH2COO – they both effectively represent the same species as CH3 CH2COONa is a strong electrolyte and completely dissociates into ions in aqueous solutions.

The only missing part in the equation is the pKa which we calculate by the negative log of Ka:

pKa = -log Ka = -log 1.32 x 10-5= 4.88

Therefore, the pH is:

${\rm{pH}}\; = \;4.88\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.75}}}}{{{\rm{0}}{\rm{.6}}}}\; = \,4.98$

The pH > pKa, and this makes sense because [HA] < [A]:

8.

What is the pH of a buffer solution that is 0.85 M pyridine (C5H5N) and 1.3 M pyridinium chloride (C5H5NHCl)? Kb (C5H5N) = 1.7 x 10-9 16.82

5.0

Solution

Pyridine is the base and the pyridinium ion produced from the dissociation of pyridinium chloride is its conjugate acid.

C5H5NHCl(aq) → C5H5NH+(aq)  + Cl(aq)

Therefore, the Henderson–Hasselbalch Equation can be shown as:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{\rm{5}}}{\rm{N]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}}}$

We have the concentration of both components of the buffer, and the ammonium ion, so the only missing part is the pKa which can be calculated by fist determining the Ka from Kb since Ka · Kb = Kw,

${K_{\rm{a}}}\,{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{b}}}}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{1.7\, \times \,{{10}^{ – 9}}}}\, = \,5.9\, \times \,{10^{ – 6}}$

Next, we calculate the pKa:

pKa = -log Ka = -log 5.9 x 10-6 = 5.2

Once the pKa is known, plug the values in the Henderson–Hasselbalch Equation:

${\rm{pH}}\; = \;5.2\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.85}}\,M}}{{{\rm{1}}{\rm{.3}}\,M}}\, = \,\,5.0$

Always check if your answer is reasonable by referring to the pH and pKa relationship:

We have pH < pKa, and this is correct because [C5H5NH+] > [C5H5N].

9.

Calculate the pH of a solution that is 0.8 M HF and 2.0 M NaF. Ka (HF) = 6.6 x 10-4

3.57

Solution

The Henderson–Hasselbalch Equation for this example is:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}$

The only missing part in the equation is the pKa which we calculate by the negative log of Ka:

pKa = -log Ka = -log 6.60 x 10-4 = 3.18

Therefore, the pH is:

${\rm{pH}}\; = \;3.18\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{2}}{\rm{.0}}}}{{{\rm{0}}{\rm{.8}}}}\; = \,3.57$

The pH > pKa, and this makes sense because [A] > [HA]:

10.

Calculate the ratio of NaNO2 to HNO2 required to create a buffer with pH = 4.00. Ka (HNO2 = 7.2 x 10-4)

7.2

Solution

The Henderson–Hasselbalch Equation for this example is:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}$

We need to determine the log term, and the only missing part in the equation is the pKa which we calculate by the negative log of Ka:

pKa = -log Ka = -log 7.2 x 10-4 = 3.14

Therefore, the pH is:

${\rm{4}}{\rm{.00}}\; = \;3.14\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}$

${\rm{log}}\frac{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\; = {\rm{4}}{\rm{.00}}\; – \,3.14\;{\rm{ = }}\;{\rm{0}}{\rm{.86}}$

$\frac{{{\rm{[N}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\; = {10^{{\rm{0}}{\rm{.86}}}}\; = \;7.2$

11.

Assuming no volume change, how many grams of sodium benzoate (C7H5O2Na) needs to be added to 250.0 mL of a 0.25 M benzoic acid (C7H5O2H) solution to prepare a buffer with a pH of 4.60? Ka = 6.46 x 10-5

23.1 g

Solution

To determine the mass of sodium benzoate, we will use the Henderson–Hasselbalch Equation to calculate its concentration first. Once the concentration is known, we can convert to moles and then to mass.

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}$

pKa = -log Ka = -log 6.46 x 10-5 = 4.19

${\rm{4}}{\rm{.60}}\; = \;4.19\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{0}}{\rm{.25}}}}$

${\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{0}}{\rm{.25}}}}\; = {\rm{4}}{\rm{.60}}\; – \;4.19\;{\rm{ = 0}}{\rm{.41}}\;$

$\frac{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{0}}{\rm{.25}}}}\; = {10^{{\rm{0}}{\rm{.41}}}}\;$

$\frac{{{\rm{[}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{0}}{\rm{.25}}}}\; = {10^{{\rm{0}}{\rm{.41}}}}\;$

[C7H5O2] = 0.64 M

Next, convert the molarity to moles, and finally to mass:

${\rm{m}}\;{\rm{(}}{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{\rm{5}}}{{\rm{O}}_{\rm{2}}}{\rm{Na)}}\;{\rm{ = }}\,{\rm{0}}{\rm{.64}}\,\frac{{\cancel{{{\rm{mol}}}}}}{{\cancel{{\rm{L}}}}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.250}}\,\cancel{{\rm{L}}}\,{\rm{ \times }}\,\frac{{{\rm{144}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}}}}}\; = \,23.1\,g$

12.

How many mL of 0.60 M HF and 0.70 M NaF must be mixed to prepare 1.00 L of a buffer solution at pH of 4.2? pKa HF = 3.8

0.18 L of 0.60 HF and 0.682 L of 0.70 NaF

Solution

There are a few ways to solve this problem, and we will show two here. The first will be a longer process, but it may help understand the key concepts. Go over both, and see which one works better for you.

Approach 1.

Let’s first determine the ratio the components using the Henderson–Hasselbalch Equation:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}$

${\rm{4}}{\rm{.2}}\; = \;3.8\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}$

${\rm{log}}\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}\; = \;0.4$

$\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}\; = \;{10^{0.4}}\; = \,2.5$

${\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}\; = \,2.5\;{\rm{[HF]}}$

Next, we set up two equations, one for the dilution of each component (M1V1 = M2V2)

Let’s say we take x L of HF, and therefore, (1 -x) L of NaF solution must be taken to make a total 1 L.

The dilution equation for HF will be:

(0.60)x = [HF] · 1L

For F, it is:

(0.70)(1 – x) = [F] · 1L

Let’s replace [F] with 2.5[HF] in the second equation, and remove the 1 L term from both equations:

(0.70) (1 – x) = 2.5[HF]

We have two equations with two unknowns:

(0.70) (1 – x) = 2.5[HF]

(0.60)x = [HF]

Now, replace the term HF in the first equation with (0.60)x:

(0.70) (1 – x) = 2.5(0.60)x

0.7 – 0.7x = 1.5x

x = 0.318 L

Therefore, we need 0.18 L of 0.60 M HF mixed with 1 – 0.318 = 0.682 L 0.70 M NaF.

Now, let’s check the calculations.

First, we are adding more base with a higher concentration than acid, and therefore, the pH is expected to be higher than the pKa. This matches the requirement because the pH must 4.2, and the pKa is 3.8.

Second, let’s calculate the concentration of each component in the final solution and plug the values in the Henderson–Hasselbalch Equation to see if they work.

M1V1 = M2V2

${{\rm{M}}_{\rm{2}}}\; = \;\frac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\;$

V2 is 1 L, so we can remove it from the equation:

(HF) M2 = M1V1 = 0.60 x 0.318 = 0.191 M

(NaF) M2 = M1V1 = 0.70 x 0.682 = 0.477 M

${\rm{pH}}\; = \;3.8\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[0}}{\rm{.477]}}}}{{{\rm{[0}}{\rm{.191]}}}}\; = \;4.197\; \approx \,4.2$

Approach 2 – a Shorter Route

We still need to determine the ratio of the components which we already know is:

${\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}\; = \,2.5\;{\rm{[HF]}}$

Now, the ratio of the concentrations and moles is the same regardless of the solution volume because converting the molarity to moles, we multiply by the same value of the volume. Therefore, we know that n (F) = 2.5 n (HF), where n represents the moles.

The second important piece of information to understand is that the moles of the components are the same in their stock solution taken to prepare the buffer, and in the buffer solution. Remember, in the basis of the M1V1 = M2Vequation is the fact that the moles taken from the stock solution are the same as they are in the final solution.

At this point, we can write two expressions for the moles of HF and NaF. Again, let’s say we take x L of HF, and therefore, (1 -x) L of NaF solution must be taken to make a total 1 L.

Using the formula for the moles; n = MV, we can write that:

n (HF) = (0.60)x

n (NaF) = (0.70)(1 – x)

So, the ratio of the moles will be:

$\frac{{{\rm{n }}\left( {{\rm{NaF}}} \right)}}{{{\rm{n }}\left( {{\rm{HF}}} \right)}}\;{\rm{ = }}\,\frac{{\left( {{\rm{0}}{\rm{.70}}} \right)\left( {{\rm{1 — x}}} \right)}}{{{\rm{0}}{\rm{.6 x}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.5}}$

1.5x = 0.7 – 0.7x

x = 0.318

As expected, both approaches give the same result. Moreover, they yield the same equation.

13.

What is the concentration of C6H5NH3Cl in the buffer solution with a pH of 4.3 containing 0.50 M C6H5NH2? Kb (C6H5NH2) 3.8 x 10-10

[C6H5NH3Cl] = 1 M

Solution

Aniline (C6H5NH2) is the base and C6H5NH3+, formed by the dissociation of the chloride salt, is its conjugate acid.

C6H5NH3Cl(aq) → C6H5NH3+(aq)  + Cl(aq)

Therefore, the Henderson–Hasselbalch Equation can be shown as:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}^{\rm{ + }}{\rm{]}}}}$

We have the concentration of both components of the buffer, and the ammonium ion, so the only missing part is the pKa which can be calculated by fist determining the Ka from Kb since Ka · Kb = Kw,

${K_{\rm{a}}}\,{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{b}}}}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{3.8\, \times \,{{10}^{ – 10}}}}\, = \,2.6\, \times \,{10^{ – 5}}$

Next, we calculate the pKa:

pKa = -log Ka = -log 2.6 x 10-5 = 4.6

Once the pKa is known, plug the values in the Henderson–Hasselbalch Equation:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}_{\rm{3}}}^{\rm{ + }}{\rm{]}}}}$

${\rm{4}}{\rm{.3}}\;{\rm{ = }}\;{\rm{4}}{\rm{.6}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.50}}\,M}}{{{\rm{x}}\;M}}$

${\rm{log}}\frac{{{\rm{0}}{\rm{.50}}\,}}{{\rm{x}}}\;{\rm{ = }}\; – {\rm{0}}{\rm{.3}}$

$\frac{{{\rm{0}}{\rm{.50}}\,}}{{\rm{x}}}\;{\rm{ = }}\;{10^ – }^{{\rm{0}}{\rm{.3}}}\; = 0.50$

0.5x = 0.50

x = 1.0

And this is the concentration of C6H5NH3Cl; [C6H5NH3Cl] = 1 M. Let’s see if this is a reasonable answer.

There is more acid than base in the solution, and therefore, pH < pKa, which matches the pH and pKa relationship:

You can always plugin the numbers to double check the calculations:

${\rm{4}}{\rm{.3}}\;{\rm{ = }}\;{\rm{4}}{\rm{.6}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.50}}\,M}}{{{\rm{0}}{\rm{.1}}\;M}}$

14.

Which of the following pairs is the best choice to prepare a buffer with pH = 3.2 ?

In what mass ratio would you add the components to prepare the buffer.

a) CH3COOH and CH3COONa b) HNO2 and KNO2 c) NH3 and NH4Cl d) C5H5N and C5H5NHCl.

$\frac{{{\rm{m}}\;{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{\;{\rm{m}}\,{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.95}}$

Solution

The best choice would be when the pKa of the acid is closest to the pH.

pKa (CH3COOH) = 4.75, pKa (HNO2) = 3.39, pKa (NH4Cl) = 9.25, pKa (C5H5NHCl) = 5.25

Therefore, the HNO2/ NaNO2 pair will be the best option to prepare a buffer with a pH of 3.2

To find the correct mass ratio, we still need to use the Henderson–Hasselbalch equation to calculate their molarity/mole ratio. So, the unknown this time is going to be the log term.

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}$

${\rm{pH}}\;{\rm{ = }}\;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}$

${\rm{3}}{\rm{.2}}\; = \;3.39\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}$

${\rm{log}}\frac{{{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\; = \; – 0.19$

$\frac{{{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\; = \;{10^{ – 0.19}}\; = \;0.65$

[NaNO2] = 0.65 [HNO2]

The ratio of molarities is equal to the mole ratio, and therefore, if we take, for example, 1 mol of HNO2, 0.65 mol of NaNO2 will be needed.

To determine the mass ratio, we multiply the moles by the corresponding molar masses:

$\frac{{{\rm{m}}\;{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{\;{\rm{m}}\,{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.65}}\,\cancel{{{\rm{mol}}}}\,{\rm{ \times }}\,\frac{{{\rm{69}}{\rm{.0 g}}\;{\rm{[NaN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}}}}}}}{{\;{\rm{1}}\,\cancel{{{\rm{mol}}}}\,{\rm{ \times }}\,\frac{{{\rm{47}}{\rm{.0 g}}\;{\rm{[HN}}{{\rm{O}}_{\rm{2}}}{\rm{]}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}}}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.95}}$

15.

What would be the pH of a buffer solution composed of 0.60 M NH3/0.45 M NH4Cl after 0.010 mole of gaseous HCl is added to a 400.0 mL solution?

9.33

Solution

The HCl is going to react with ammonia thus consuming some of the base:

HCl(aq) + NH3(aq) → NH4Cl(aq)

Let’s first convert the moles of HCl to concentration and do the stoichiometric calculations with molarity.

M (HCl) = 0.010 mol/0.4000 L = 0.025 M

Because the HCl is the limiting reactant, therefore, there will be 0.025 M of NH4Cl produced in this reaction according to their 1:1 molar ratio with. The concentration of ammonia, on the other hand, will decrease by 0.025 M.

NH3(aq) + HCl(aq) → NH4Cl(aq)

Notice that we put approximately zero for the concentration of H+ even though the HCl reacts completely with ammonia. There is always some H+ in the solution without which we couldn’t calculate the pH, and this H+ comes from the hydrolysis of the ammonium ion.

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}$

${\rm{pH}}\; = \;9.25\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[0}}{\rm{.575]}}}}{{{\rm{[0}}{\rm{.475]}}}}\; = \;9.33$

16.

Calculate the pH after 20.0 g NaOH(s) is added to 1.0 L of a buffer solution containing 1.80 M acetic acid and 1.50 M sodium acetate at pH 4.0.

4.93

Solution

First, calculate the concentration of NaOH:

${\rm{M}}\;{\rm{(NaOH)}}\; = \;{\rm{2}}{\rm{.00}}\,{\rm{g}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{40}}{\rm{.0}}\,{\rm{g}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,}}{{{\rm{1}}\,{\rm{L}}}}\; = \;0.0500\,M$

This amount of sodium hydroxide is going to react with the acetic acid, decreasing its concentration to 1.80 – 0.500 = 1.3 M.

As a result of this reaction, there is more acetate ion formed, and therefore, we need to find its final concentration by adding the change to the initial concentration.

CH3CO2H(aq) + NaOH(aq) → CH3CO2Na(aq) + H2O(aq)

The pH is then calculated by entering all the numbers in the equation:

${\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH]}}}}$

${\rm{pH}}\; = \;4.75\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{2}}{\rm{.0}}}}{{{\rm{1}}{\rm{.3}}}}\; = \,4.93$