Buffer Solutions Practice Problems

Acids and Bases

This is a summary practice problem set on buffer solutions aimed to help identify buffers, calculating the pH of a buffer solution prepared from a weak acid and its conjugate base or vice versa.

The links to the corresponding topics are given below:

Practice

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.

answer

(d) Na₂HPO₄ + NaH₂PO₄, (e) CH₃CH₂NH₂ + CH₃CH₂NH₃⁺ (f) NaNO₂ + HNO₂, (g) KCN + HCN, (h) Na₂SO₄ + NaHSO_4, (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.

(d) Na₂HPO₄is a weak base (HPO₄^2- is the conjugate base of a weak acid H₂PO₄^–) and NaH₂PO₄is 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) HNO₂is a weak acid and NO₂^–is its conjugate base, therefore, this is a buffer solution.

(g) HCN is a weak acid and CN^–is 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) NH₃is 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.

Addition of which salt will suppress the ionization of HOCl?

Na₂SO₄
KNO₃
NaOCl
NaNO₂

answer

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.

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.

answer

NH₃(aq) + HNO₃(aq) → NH₄HNO₃(aq)

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

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)

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

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

answer

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 NH_3, 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 pK_a_.

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

NH₃(aq) + H₂O(aq) ⇆ NH₄⁺(aq) + OH^–(aq) K_b (NH₃) = 1.8 x 10^-5

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

The pK_acan be calculated by either first determining the K_afrom K_b since K_a· K_b= K_w_,or by first determining the pK_b and converting using the relationship: pK_a + pK_b = 14

Let’s go with the second option this time:

pK_b= -log K_b= -log 1.8 x 10^-5= 4.75

Therefore, the pK_a is:

pK_a = 14 – pK_b = 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 pK_a 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 K_b, so let’s write the reaction of ammonia with water:

NH₃(aq) + H₂O(aq) ⇆ NH₄⁺(aq) + OH^–(aq) K_b (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 NH₃is 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

	[NH₃]	[NH₄⁺]	[OH^–]
Initial	0.75	0.95	0
Change	-x	+x	+x
Equil	0.75 – x	0.95 + x	x

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

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^-5and 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.

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

answer

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 A^–is 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 pK_a which we calculate by the negative log of K_a:

pK_a= -log K_a = -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^–]:

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

answer

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 pK_a which we calculate by the negative log of K_a:

pK_a= -log K_a = -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]:

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

answer

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 pK_a which we calculate by the negative log of K_a:

pK_a= -log K_a = -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^–]:

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

answer

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 pK_awhich can be calculated by fist determining the K_afrom K_b since K_a· K_b= K_w_,

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

Next, we calculate the pK_a:

pK_a= -log K_a= -log 5.9 x 10^-6= 5.2

Once the pK_ais 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 pK_a relationship:

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

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

answer

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 pK_a which we calculate by the negative log of K_a:

pK_a= -log K_a = -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. K_a (HNO₂ = 7.2 x 10^-4)

answer

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 pK_a which we calculate by the negative log of K_a:

pK_a= -log K_a = -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? K_a = 6.46 x 10^-5

answer

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.

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

pK_a= -log K_a = -log 6.46 x 10^-5 = 4.19

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

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

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

[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? pK_a HF = 3.8

answer

0.18 L of 0.60 M HF and 0.682 L of 0.70 M 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 pK_a.This matches the requirement because the pH must 4.2, and the pK_ais 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) M_{2 =}M₁V₁= 0.60 x 0.318 = 0.191 M

(NaF) M_{2 =}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 = M₂V₂equation 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? K_b (C₆H₅NH₂) 3.8 x 10^-10

answer

[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:

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

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

Next, we calculate the pK_a:

pK_a= -log K_a= -log 2.6 x 10^-5= 4.6

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

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

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

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

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 < pK_a, which matches the pH and pK_a 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.

answer

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

Solution

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

pK_a(CH₃COOH) = 4.75, pK_a(HNO₂) = 3.39, pK_a(NH₄Cl) = 9.25, pK_a(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?

answer

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)

	[ NH₃]	[H⁺]	[NH₄⁺]
Before Reaction	0.60	0.025	0.45
Change	-0.025	-0.025	+0.025
After Reaction	0.575	≈0	0.475

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.

answer

4.93

Solution

First, calculate the concentration of NaOH:

\[{\rm{M}}\;{\rm{(NaOH)}}\; = \;{\rm{20}}{\rm{.0}}\,{\rm{g}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{40}}{\rm{.0}}\,{\rm{g}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,}}{{{\rm{1}}\,{\rm{L}}}}\; = \;0.500\,M\]

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)

	[ CH₃CO₂H]	[ NaOH]	[CH₃CO₂^–]
Before Reaction	1.80	0.50	1.50
Change	-0.500	-0.500	+0.500
After Reaction	2.30	≈0	2.00

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

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

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

17.

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General Chemistry

Acids and Bases

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