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Chemical transformations · Acids and bases

Extension: buffer solutions

Your blood stays between pH 7.35 and 7.45 even though your cells are constantly producing acid, for example when you sprint down a basketball court. A drift of just a few tenths of a pH unit would stop your proteins working properly. The control is done by a surprisingly simple chemical system: a buffer solution. By the end of this page you will be able to explain how it works and where it runs out.

What a buffer is and how it works

Add 1 mL of 0.1 M HCl to 100 mL of pure water and the pH collapses from 7 to about 3. Add the same milliliter to 100 mL of a buffer containing 0.1 M acetic acid and 0.1 M sodium acetate and the pH hardly moves, from 4.76 to about 4.75. Why can some solutions soak up so much acid or base?

Diagram of how a buffer works: HA neutralizes added base and A⁻ neutralizes added acid HA / A⁻ in similar amounts + strong base HA neutralizes it + strong acid A⁻ neutralizes it

The two reactions are:

  • Added acid: $\mathrm{A^- + H^+ \rightarrow HA}$

  • Added base: $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$

Because both HA and A⁻ are present in quantity, neither addition uses up the "reserve" at once.

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A buffer solution consists of a weak acid or base and its conjugate species in similar concentrations. At $25\,\mathrm{^\circ C}$, three aqueous solutions are prepared…

A buffer solution consists of a weak acid or base and its conjugate species in similar concentrations. At $25\,\mathrm{^\circ C}$, three aqueous solutions are prepared by mixing equal volumes of the following pairs of solutes, each at an initial concentration of $0.10\,\mathrm{mol\times L^{-1}}$:

1) $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COONa}$

2) $\mathrm{HCl}$ and $\mathrm{NaCl}$

3) $\mathrm{NH_3}$ and $\mathrm{NH_4Cl}$

  1. a) Justify, with reasoning, which of these three mixtures constitute a buffer solution.
  2. b) Explain which component of each buffer mixture neutralizes a small amount of strong acid and which neutralizes a small amount of strong base.
Solve

The Henderson–Hasselbalch equation

How the pH of a buffer compares with the pKa depending on the ratio of conjugate base to weak acid pH = pKₐ + log([A⁻]/[HA]) [A⁻] = [HA] pH = pKₐ [A⁻] > [HA] pH > pKₐ [A⁻] < [HA] pH < pKₐ
Practice★☆☆Not started
The Henderson-Hasselbalch equation for a conjugate acid-base pair $\mathrm{HA/A^-}$ is written as: $$\mathrm{pH} = \mathrm{p}Ka +…

The Henderson-Hasselbalch equation for a conjugate acid-base pair $\mathrm{HA/A^-}$ is written as:

$$\mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)$$

At $25\,\mathrm{^\circ C}$ there is a buffer made up of benzoic acid ($\mathrm{C_6H_5COOH}$, $K_a = 6.3\times 10^{-5}$) and sodium benzoate ($\mathrm{C_6H_5COONa}$).

  1. a) State two fundamental hypotheses or assumptions that justify using the analytical concentrations of the prepared components in this equation as an approximation to the equilibrium concentrations.
  2. b) If a $\mathrm{pH} = 4.50$ is to be set, calculate the molar ratio $[\mathrm{A^-}]/[\mathrm{HA}]$ at equilibrium at that temperature.
Solve
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$250\,\mathrm{mL}$ of a buffer solution containing acetic acid ($\mathrm{CH_3COOH}$) $0.20\,\mathrm{mol\times L^{-1}}$ and sodium acetate ($\mathrm{CH_3COONa}$)…

$250\,\mathrm{mL}$ of a buffer solution containing acetic acid ($\mathrm{CH_3COOH}$) $0.20\,\mathrm{mol\times L^{-1}}$ and sodium acetate ($\mathrm{CH_3COONa}$) $0.30\,\mathrm{mol\times L^{-1}}$ is available. To this solution, $0.010\,\mathrm{mol}$ of solid sodium hydroxide ($\mathrm{NaOH}$) is added, with no appreciable change in the final volume. Data at $25\,\mathrm{^\circ C}$: $K_a(\mathrm{CH_3COOH}) = 1.8\times 10^{-5}$.

  1. a) Write the chemical neutralization reaction that takes place first and calculate the amounts of acetic acid and acetate ion present after neutralization.
  2. b) Calculate the $\mathrm{pH}$ of the resulting solution at $25\,\mathrm{^\circ C}$ using the Henderson-Hasselbalch equation. For buffers, use the approximation $\mathrm{pH}=\mathrm{p}K_a+\log(n_{\mathrm{base}}/n_{\mathrm{ácido}})$ after stoichiometric neutralization. For the ammonium/ammonia pair, $K_a=K_w/K_b$. If one component is used up, solve the equilibrium of the remaining species. Assume ideal behavior and additive volumes.
Solve
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A basic buffer solution is prepared by dissolving ammonia ($\mathrm{NH_3}$) and ammonium chloride ($\mathrm{NH_4Cl}$) in water at $25\,\mathrm{^\circ C}$, so that the…

A basic buffer solution is prepared by dissolving ammonia ($\mathrm{NH_3}$) and ammonium chloride ($\mathrm{NH_4Cl}$) in water at $25\,\mathrm{^\circ C}$, so that the initial concentrations are $[\mathrm{NH_3}] = 0.15\,\mathrm{mol\times L^{-1}}$ and $[\mathrm{NH_4^+}] = 0.10\,\mathrm{mol\times L^{-1}}$. To $500\,\mathrm{mL}$ of this mixture, $5.0\,\mathrm{mL}$ of concentrated hydrochloric acid $1.0\,\mathrm{mol\times L^{-1}}$ is added. Data at $25\,\mathrm{^\circ C}$: $K_b(\mathrm{NH_3}) = 1.8\times 10^{-5}$, $K_w = 1.0\times 10^{-14}$.

  1. a) Calculate the initial $\mathrm{pH}$ of the buffer solution.
  2. b) Set up the material balance after the addition of the strong acid and determine the new value of $\mathrm{pH}$ once chemical equilibrium has been reached. For buffers, use the approximation $\mathrm{pH}=\mathrm{p}K_a+\log(n_{\mathrm{base}}/n_{\mathrm{ácido}})$ after stoichiometric neutralization. For the ammonium/ammonia pair, $K_a=K_w/K_b$. If one component is used up, solve the equilibrium of the remaining species. Assume ideal behavior and additive volumes.
Solve

Buffer capacity and its limits

Add NaOH slowly to an acetic acid/acetate buffer and the pH creeps up almost unnoticeably for many millilitres. Then, once the acid component is used up, the pH rises sharply with each further drop. What limit makes the buffering collapse so suddenly?

Curve of pH against volume of strong base added, showing the buffering plateau and the sharp rise when the capacity is exceeded buffered zone capacity exhausted 0V(eq) volume of base added pH
Practice★★☆Not started
In the lab, the resistance to change in $\mathrm{pH}$ of two solutions of equal volume ($100\,\mathrm{mL}$) at $25\,^\circ\mathrm{C}$ is compared: beaker A contains pure…

In the lab, the resistance to change in $\mathrm{pH}$ of two solutions of equal volume ($100\,\mathrm{mL}$) at $25\,^\circ\mathrm{C}$ is compared: beaker A contains pure deionized water and beaker B contains an acetic acid / sodium acetate buffer $0.10\,\mathrm{mol\times L^{-1}}$ of each component ($K_a = 1.75\times 10^{-5}$, $\mathrm{p}K_a = 4.76$). The same dose of $1.0\,\mathrm{mL}$ of $\mathrm{HCl}$ $0.10\,\mathrm{mol\times L^{-1}}$ is added to both beakers. The figure summarizes the experiment and the $\mathrm{pH}$ values recorded before and after the addition.

Comparison of the buffering effect upon addition of hydrochloric acid + 1.0 mL HCl 0.10 M + 1.0 mL HCl 0.10 M 100 mL Beaker A: Pure water 100 mL Beaker B: HA/A⁻ buffer (0.10 M) Beaker Initial pH pH after addition ΔpH Pure water 7.00 3.00 −4.00 Acetic/acetate buffer 4.76 4.75 −0.01
  1. a) Explain the difference in the change in $\mathrm{pH}$ between beaker A ($\Delta\mathrm{pH} = -4.00$) and beaker B ($\Delta\mathrm{pH} = -0.01$), deducing the reactions involved.
  2. b) Define the concept of finite buffer capacity and state what would happen qualitatively to the $\mathrm{pH}$ of beaker B if $150\,\mathrm{mL}$ of $\mathrm{HCl}$ $0.10\,\mathrm{mol\times L^{-1}}$ were added.
Solve
Practice★★★Not started
You have $200\,\mathrm{mL}$ of a buffer made up of methanoic acid (formic acid, $\mathrm{HCOOH}$, $K_a = 1.8\times 10^{-4}$) $0.12\,\mathrm{mol\times L^{-1}}$ and sodium…

You have $200\,\mathrm{mL}$ of a buffer made up of methanoic acid (formic acid, $\mathrm{HCOOH}$, $K_a = 1.8\times 10^{-4}$) $0.12\,\mathrm{mol\times L^{-1}}$ and sodium methanoate ($\mathrm{HCOONa}$) $0.080\,\mathrm{mol\times L^{-1}}$ at $25\,\mathrm{^\circ C}$. The useful buffering capacity is considered exhausted when the $\mathrm{pH}$ falls below $\mathrm{p}K_a - 1.00$.

  1. a) Calculate the volume of $\mathrm{HCl}$ $0.50\,\mathrm{mol\times L^{-1}}$ solution that must be added to reach approximately the lower limit of the buffering capacity ($\mathrm{pH} = \mathrm{p}K_a - 1.00$).
  2. b) Calculate the $\mathrm{pH}$ that would be obtained if the exact stoichiometric amount of $\mathrm{HCl}$ needed to neutralize all the methanoate ion initially present were added, assuming additive volumes. For buffers, use the approximation $\mathrm{pH}=\mathrm{p}K_a+\log(n_{\mathrm{base}}/n_{\mathrm{ácido}})$ after stoichiometric neutralization. For the ammonium/ammonia pair, $K_a=K_w/K_b$. If one component is used up, solve the equilibrium of the remaining species. Assume ideal behavior and additive volumes.
Solve

Buffers in the body and the laboratory

Diagram of the bicarbonate buffer in the body, with the lungs controlling CO₂ and the kidneys controlling HCO₃⁻ blood pH 7.35–7.45 lungs control CO₂ kidneys control HCO₃⁻
Practice★☆☆Not started
Human blood plasma contains a key physiological buffer system made up of carbonic acid and the bicarbonate ion: $$\mathrm{CO_2(ac)} + \mathrm{H_2O(l)} \rightleftharpoons…

Human blood plasma contains a key physiological buffer system made up of carbonic acid and the bicarbonate ion:

$$\mathrm{CO_2(ac)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2CO_3(ac)} \rightleftharpoons \mathrm{HCO_3^-(ac)} + \mathrm{H^+(ac)}$$

At body temperature, the effective value of $\mathrm{p}K_a$ for the bicarbonate/total dissolved CO₂ system is $6.10$, and the $\mathrm{pH}$ of the plasma is maintained at around $7.40$ under normal conditions.

  1. a) Calculate the ratio $[\mathrm{HCO_3^-}]/[\mathrm{CO_2^*}]$ in blood plasma at that $\mathrm{pH}$.
  2. b) Describe chemically how this buffer pair responds to an increase in the concentration of protons from cellular metabolism, preventing sudden drops in $\mathrm{pH}$. Here $[\mathrm{CO_2^*}]=[\mathrm{CO_2(ac)}]+[\mathrm{H_2CO_3}]$ and the effective relationship $\mathrm{pH}=6.10+\log([\mathrm{HCO_3^-}]/[\mathrm{CO_2^*}])$ is used; total dissolved CO₂ is not identified with the concentration of molecular carbonic acid.
Solve

Summary

See all the exercises in this section