AP Chemistry 8.7-8.10: pKa, Henderson-Hasselbalch, and Buffer Capacity

Buffer solution

A solution containing appreciable amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid. It resists large pH changes when small amounts of strong acid or strong base are added.

What species make up an acetic acid–acetate buffer?

The weak acid is acetic acid, $\mathrm{CH_3COOH}$, and its conjugate base is acetate, $\mathrm{CH_3COO^-}$, supplied by a soluble salt such as sodium acetate, $\mathrm{CH_3COONa}$.

What species make up an ammonia–ammonium buffer?

The weak base is ammonia, $\mathrm{NH_3}$, and its conjugate acid is ammonium, $\mathrm{NH_4^+}$, supplied by a salt such as ammonium chloride, $\mathrm{NH_4Cl}$.

What equilibrium explains the buffering action of an acetic acid–acetate solution?

$\mathrm{CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-}$. Added base removes $\mathrm{H_3O^+}$ and shifts the equilibrium right, whereas added acid increases $\mathrm{H_3O^+}$ and shifts it left.

How does a buffer respond when a small amount of strong acid is added?

The conjugate base consumes the added hydronium ions: $\mathrm{A^- + H_3O^+ \rightarrow HA + H_2O}$. This converts strong acid into the weak acid of the buffer pair, so the pH decreases only slightly.

How does a buffer respond when a small amount of strong base is added?

The weak acid consumes the added hydroxide ions: $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$. This converts strong base into the weak conjugate base, so the pH increases only slightly.

Why does a buffer resist pH changes more effectively than an unbuffered solution?

The buffer components react nearly stoichiometrically with added $\mathrm{H_3O^+}$ or $\mathrm{OH^-}$, converting them into weak acid or weak base species. These products ionize only partially, causing a much smaller change in $[\mathrm{H_3O^+}]$.

What is the acid dissociation expression for a weak acid buffer pair $\mathrm{HA/A^-}$?

$K_a=\dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$. Rearranging gives $[\mathrm{H_3O^+}]=K_a\dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]}$.

What does $\mathrm{p}K_a$ represent, and how is it related to $K_a$?

$\mathrm{p}K_a=-\log K_a$. A smaller $\mathrm{p}K_a$ corresponds to a larger $K_a$ and therefore a stronger acid.

Henderson–Hasselbalch equation

For a weak acid and its conjugate base, $\mathrm{pH}=\mathrm{p}K_a+\log\left(\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)$. It estimates buffer pH when both components are present in appreciable amounts and the weak-acid equilibrium change is small relative to their initial concentrations.

What is the pH of a buffer when $[\mathrm{A^-}]=[\mathrm{HA}]$?

The Henderson–Hasselbalch equation gives $\mathrm{pH}=\mathrm{p}K_a$ because $\log(1)=0$.

How does the conjugate-base-to-weak-acid ratio affect buffer pH?

For $\mathrm{pH}=\mathrm{p}K_a+\log([\mathrm{A^-}]/[\mathrm{HA}])$, increasing $[\mathrm{A^-}]/[\mathrm{HA}]$ raises pH, while decreasing it lowers pH. A tenfold ratio corresponds to a pH one unit above $\mathrm{p}K_a$; a one-tenth ratio corresponds to a pH one unit below $\mathrm{p}K_a$.

What assumptions justify using the Henderson–Hasselbalch equation for a weak-acid buffer?

The acid is treated as a monoprotic weak acid, water autoionization is negligible, the conjugate-base salt is completely dissociated, and the equilibrium change $x$ is small compared with the initial buffer concentrations. Concentration ratios are also used as approximations to activity ratios.

When is the Henderson–Hasselbalch equation generally not reliable?

It becomes unreliable when one buffer component is nearly absent, when concentrations are very low so water autoionization matters, or when the weak-acid equilibrium change is not small. It is also not directly applicable to overlapping equilibria of a polyprotic acid unless the relevant $\mathrm{p}K_a$ values are sufficiently separated.

How can the Henderson–Hasselbalch equation be applied to a weak-base buffer?

Treat the protonated base as the weak acid: $\mathrm{BH^+\rightleftharpoons B+H^+}$. Then $\mathrm{pH}=\mathrm{p}K_a(\mathrm{BH^+})+\log([\mathrm{B}]/[\mathrm{BH^+}])$, where $\mathrm{p}K_a(\mathrm{BH^+})=\mathrm{p}K_w-\mathrm{p}K_b$.

What is the relationship between the $\mathrm{p}K_a$ of a conjugate acid and the $\mathrm{p}K_b$ of its base at $25\ ^\circ\mathrm{C}$?

$\mathrm{p}K_a+\mathrm{p}K_b=\mathrm{p}K_w\approx14.00$, so $\mathrm{p}K_a\approx14.00-\mathrm{p}K_b$.

How should the pH of a buffer be calculated after adding a strong acid or strong base?

First use stoichiometry to determine how many moles of $\mathrm{HA}$ and $\mathrm{A^-}$ remain or are formed. Then divide by the total volume if concentrations are needed and apply the Henderson–Hasselbalch equation. The neutralization reaction should be completed before the equilibrium calculation.

Why are mole ratios often used instead of concentration ratios in buffer calculations after mixing?

Both buffer components occupy the same final solution volume, so $[\mathrm{A^-}]/[\mathrm{HA}]$ equals $n(\mathrm{A^-})/n(\mathrm{HA})$. Thus, $\mathrm{pH}=\mathrm{p}K_a+\log\left(n_{\mathrm{A^-}}/n_{\mathrm{HA}}\right)$ can be used directly after stoichiometric neutralization, provided the assumptions remain valid.

A buffer initially contains equal concentrations of acetic acid and acetate. What happens when a small amount of $\mathrm{NaOH}$ is added?

The hydroxide converts some $\mathrm{CH_3COOH}$ into $\mathrm{CH_3COO^-}$. Therefore, the acid amount decreases and the base amount increases, causing a small pH increase rather than the very large increase expected in an unbuffered solution.

A solution contains $0.0100$ mol each of $\mathrm{HA}$ and $\mathrm{A^-}$. How does adding $0.00100$ mol $\mathrm{OH^-}$ change the buffer composition?

The reaction $\mathrm{HA+OH^-\rightarrow A^-+H_2O}$ leaves $0.00900$ mol $\mathrm{HA}$ and produces a total of $0.01100$ mol $\mathrm{A^-}$. The pH is then $\mathrm{p}K_a+\log(0.01100/0.00900)$.

What is the correct sequence for solving a strong-acid addition problem involving a buffer?

Calculate moles of added strong acid, use $\mathrm{A^-+H_3O^+\rightarrow HA+H_2O}$ to update the buffer moles, then calculate the pH from the new $\mathrm{A^-}/\mathrm{HA}$ ratio. Do not first treat the buffer as if all of its species were strong electrolytes.

Why does a small amount of strong base cause a dramatic pH increase in an unbuffered acidic solution?

The added hydroxide may exceed the small initial amount of hydronium ions. Once the hydronium is consumed, excess $\mathrm{OH^-}$ remains, and the pH is determined by its concentration through $\mathrm{pOH=-\log[OH^-]}$ and $\mathrm{pH}=14.00-\mathrm{pOH}$ at $25\ ^\circ\mathrm{C}$.

What is the correct sequence for solving a strong-base addition problem involving an unbuffered weak-acid solution?

Determine the initial moles of hydronium or weak acid as appropriate, compare them with the added hydroxide moles, and perform stoichiometric neutralization first. If excess hydroxide remains, calculate its concentration and use $\mathrm{pOH=-\log[OH^-]}$ followed by $\mathrm{pH}=14.00-\mathrm{pOH}$ at $25\ ^\circ\mathrm{C}$.

Why does adding $1.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}$ $\mathrm{NaOH}$ to $100\ \mathrm{mL}$ of a $0.10\ \mathrm{M}/0.10\ \mathrm{M}$ acetate buffer change its pH only slightly?

The added hydroxide is only $1.0\times10^{-4}$ mol, compared with $1.0\times10^{-2}$ mol of each buffer component. It converts a small fraction of acetic acid to acetate, changing the ratio only slightly and therefore changing pH from about $4.74$ to $4.75$.

What is buffer capacity?

Buffer capacity is the amount of strong acid or strong base that can be added to a specified volume of buffer before the pH changes significantly, often defined as a change of about one pH unit.

How does total buffer concentration affect buffer capacity?

At the same acid-to-base ratio, a buffer with greater concentrations of both components has greater capacity because it contains more moles available to neutralize added acid or base. Total concentration affects capacity, whereas the ratio primarily determines pH.

How does buffer composition affect its capacity toward added acid versus added base?

Capacity toward added strong acid depends mainly on the amount of conjugate base available to consume it. Capacity toward added strong base depends mainly on the amount of weak acid available to neutralize it. A roughly equal mixture provides substantial capacity in both directions.

What composition gives a buffer its greatest practical effectiveness?

The weak acid and conjugate base should be present in roughly comparable amounts. This provides substantial capacity against both added acid and added base and places the pH near the buffer pair's $\mathrm{p}K_a$.

What concentration guideline indicates that a buffer is becoming ineffective?

A buffer generally loses usefulness when one component falls below about $10\%$ of the other. At that point, the pH is about one unit away from the $\mathrm{p}K_a$ and the depleted component has little remaining capacity.

Why does a buffer eventually lose its effectiveness?

Adding enough strong acid or base substantially depletes one member of the conjugate pair. Once one component is nearly exhausted, additional reagent is no longer efficiently converted into a weak species, so the pH changes rapidly.

How should a buffer pair be selected for a desired pH?

Choose a weak acid whose $\mathrm{p}K_a$ is close to the target pH, ideally within about one pH unit. Use the Henderson–Hasselbalch equation to select the required ratio: $[\mathrm{A^-}]/[\mathrm{HA}]=10^{\mathrm{pH}-\mathrm{p}K_a}$.

Which type of buffer is generally best for acidic pH values, and which for basic pH values?

Weak-acid/conjugate-base buffers are generally best for pH values below 7. Weak-base/conjugate-acid buffers are generally best for pH values above 7.

Why does dilution usually have little effect on the pH of an ideal buffer?

Dilution decreases the concentrations of both $\mathrm{HA}$ and $\mathrm{A^-}$ by approximately the same factor, leaving their ratio nearly unchanged in $\mathrm{pH}=\mathrm{p}K_a+\log([\mathrm{A^-}]/[\mathrm{HA}])$. However, dilution decreases buffer capacity because fewer moles of each component are present per volume.

How does adding a strong acid or base differ from simply diluting a buffer?

A strong acid or base changes the ratio of conjugate-pair components through neutralization, so it changes pH. Dilution generally preserves that ratio and therefore changes pH little, although it reduces the buffer's capacity.