Question

A $25.0-\mathrm{mL}$ solution of $0.100\mathrm{M}{\mathrm{CH}}_3\mathrm{COOH}$ is titrated with a $0.200\mathrm{M}\mathrm{KOH}$ solution. Calculate the $\mathrm{pH}$ after the following additions of the $\mathrm{KOH}$ solution: (a) $0.0\mathrm{mL},$ (b) $5.0\mathrm{mL}$ (c) $10.0\mathrm{mL}$ (d) $12.5\mathrm{mL}$ (e) $15.0\mathrm{mL}$

Short answer

(a) 2.87, (b) 4.56, (c) 5.34, (d) 8.72, (e) 12.10

Step-by-step solution

Initial Setup

First, identify the initial moles of acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) before any $\mathrm{KOH}$ is added. The moles of ${\mathrm{CH}}_3\mathrm{COOH}$ is given by $\text{moles}=\text{volume}\times \text{molarity}$. Hence, $0.025\mathrm{L}\times 0.100\mathrm{M}=0.0025\mathrm{moles}$.

Determine pH Before Addition of KOH (0.0 mL)

Since no base is added, the solution contains only ${\mathrm{CH}}_3\mathrm{COOH}$, a weak acid with $K_a=1.8\times {10}^{-5}$. Set up the equilibrium ${\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}$. Create the ICE table, solve for $x$ using the approximation $K_a=\frac{x^2}{0.100-x}\approx \frac{x^2}{0.100}$. Solving for $x$, $x=\sqrt{1.8\times {10}^{-6}}$, gives $x=1.34\times {10}^{-3}$. Therefore, $\text{pH}=-\log (1.34\times {10}^{-3})\approx 2.87$.

After Addition of 5.0 mL KOH

Calculate moles of $\mathrm{KOH}$ added: $0.005L\times 0.200M=0.001\text{ moles}$. Compute remaining ${\mathrm{CH}}_3\mathrm{COOH}$: $0.0025-0.001=0.0015\text{ moles}$. Form ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ at same moiety of $0.001\text{ moles}$. Use the Henderson-Hasselbalch equation: $\text{pH}=\text{p}{K}_a+\log \left(\frac{0.001}{0.0015}\right)$. With $\text{p}{K}_a=4.74$, $\text{pH}=4.74+\log (0.667)\approx 4.56$.

After Addition of 10.0 mL KOH

Calculate moles of $\mathrm{KOH}$ added: $0.01L\times 0.200M=0.002\text{ moles}$. Compute remaining ${\mathrm{CH}}_3\mathrm{COOH}$: $0.0025-0.002=0.0005\text{ moles}$. Form ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ at $0.002\text{ moles}$ with excess.Use Henderson-Hasselbalch equation: $\text{pH}=\text{p}{K}_a+\log \left(\frac{0.002}{0.0005}\right)$. With $\text{p}{K}_a=4.74$, $\text{pH}=4.74+\log (4)\approx 5.34$.

After Addition of 12.5 mL KOH (Equivalence Point)

Calculate moles of $\mathrm{KOH}$ added: $0.0125L\times 0.200M=0.0025\text{ moles}$. This is equal to initial moles of ${\mathrm{CH}}_3\mathrm{COOH}$, resulting in complete titration. At the equivalence point, concentration of $[{\mathrm{CH}}_3{\mathrm{COO}}^{-}]$ in 37.5 mL total solution is $\frac{0.0025}{0.0375}=0.0667\text{ M}$. Calculate $[{\mathrm{OH}}^{-}]$ using $K_w/K_a$: $K_b=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}$, and find ${\mathrm{OH}}^{-}$ concentration. Finally, find $\text{pH = 8.72}$.

After Addition of 15.0 mL KOH

Calculate moles of $\mathrm{KOH}$ added: $0.015L\times 0.200M=0.003\text{ moles}$. Excess ${\mathrm{OH}}^{-}$ = $0.003-0.0025=0.0005\text{ moles}$. Calculate $[{\mathrm{OH}}^{-}]$ as $\frac{0.0005}{0.040}=0.0125\text{ M}$. Calculate pOH = $-\log (0.0125)\approx 1.90$, giving $\text{pH}=14-1.90=12.10$.

Key concepts

Acid-Base Equilibrium

Acid-base equilibrium is crucial for understanding reactions between acids and bases in solutions. When a weak acid like acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) is in water, it partly dissociates into its ions: ${\mathrm{H}}^{+}$ (a proton) and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ (acetate ion). This dissociation can be expressed by an equilibrium expression reflecting the balance between the reactants and products.

• The more the acid dissociates, the stronger it is.
• Equilibrium concepts help us understand this dissociation in quantitative terms.

The equilibrium constant $K_a$ for a weak acid is defined as:$$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{CH}}_3{\mathrm{COO}}^{-}]}{[{\mathrm{CH}}_3\mathrm{COOH}]}$$Here, $K_a$ provides insight into the position of the equilibrium and the strength of the acid. A higher $K_a$ value means a stronger acid which dissociates more. In acetic acid titration, knowing the $K_a$ allows calculation of the changes in ${\mathrm{H}}^{+}$ concentration as the base is added.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simplified method for pH calculations in buffer solutions, turning the complex equilibrium dynamics into more manageable arithmetic. It assists in determining the pH during titrations involving weak acids or bases and their salts.
When a weak acid (${\mathrm{CH}}_3\mathrm{COOH}$) and its conjugate base (${\mathrm{CH}}_3{\mathrm{COO}}^{-}$) are present, you use the equation:$$pH=p{K}_a+\log \left(\frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}\right)$$

• $p{K}_a$ is the negative logarithm of the acid dissociation constant.
• The fraction $\frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}$ is the ratio of the concentration of the conjugate base to the acid.

This equation presumes that the concentrations of acid and conjugate base are similar, a condition often met in buffer solutions or pre-equivalence stages of titration. It's a powerful tool for pH prediction as it accounts for shifts in concentration due to added base or acid.

pH Calculation

Calculating pH is about determining the acidity or basicity of a solution quantitatively, using the concentration of hydrogen ions $[{\mathrm{H}}^{+}]$. During titration of acetic acid with $\mathrm{KOH}$, the pH changes at each step: initial, any point before and after equivalence, and finally when excess base is present.

• Before adding any $\mathrm{KOH}$, the pH depends on undissociated acetic acid, calculated using its $K_a$.
• The titration equivalence point is when equal moles of acid and base react, often resulting in a different pH due to the nature of the resulting solution.
• Beyond the equivalence point, pH depends on excess base.

Each addition of $\mathrm{KOH}$ changes the pH as it consumes ${\mathrm{H}}^{+}$ and increases $[{\mathrm{OH}}^{-}]$. To compute the resulting pH, conversion from $[{\mathrm{OH}}^{-}]$ to $\mathrm{pOH}$ and accordingly to pH, via $\mathrm{pH}=14-\mathrm{pOH}$, is necessary once excess ${\mathrm{OH}}^{-}$ is present. Consistently track these changes to accurately predict solution acidity or basicity.