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A 25.0 mL sample contains both hydrochloric acid and acetic acid…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A $25.0\ \text{mL}$ sample contains both hydrochloric acid and acetic acid. It is titrated with $0.100\ \text{mol L}^{-1}$ sodium hydroxide, and the conductivity of the mixture is measured after each addition. Each reading is multiplied by $(V_0 + V)/V_0$ to allow for the dilution, and the figure shows the result. It falls steeply, rises gently, and then rises steeply. Straight lines through the three parts meet at $7.50\ \text{mL}$ and at $20.0\ \text{mL}$ , the two end points. Use NCERT's limiting molar ionic conductivities, in $\text{S cm}^2\ \text{mol}^{-1}$ : $\text{H}^+$ 349.6, $\text{Na}^+$ 50.1, $\text{Cl}^-$ 76.3, $\text{CH}_3\text{COO}^-$ 40.9 and $\text{OH}^-$ 199.1. Take $K_a = 1.74 \times 10^{-5}\ \text{mol L}^{-1}$ for acetic acid. For a conductivity, let each ion add $c\lambda^\circ$ , a fair estimate in solutions this dilute. Ignore the ions from water, and the very small ionisation of acetic acid while hydrochloric acid is present. Give conductivities in S per metre.

A graph of the conductivity, multiplied by the dilution factor, against the volume of sodium hydroxide added, in millilitres, from 0 to 30 mL. The vertical axis has no numbers. The points fall steeply in a straight line, then rise gently in a second straight line, then rise steeply in a third. Dashed lines extend each part, and they meet at 7.50 mL and at 20.0 mL, where the two end points are marked.

What is the concentration of hydrochloric acid in the sample?
What is the concentration of acetic acid in the sample?
What is the pH at the first end point?
What is the measured conductivity of the sample before any base is added?
What is the measured conductivity at the second break (the reading itself, not the value corrected for dilution)?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Read the two breaks as two end points, then count the ions present at each stage.

(a) and (b). The first break is the end of the strong acid: $c(\text{HCl}) = \dfrac{0.100 \times 7.50}{25.0} = 0.0300\ \text{mol L}^{-1}$ . The next $12.5\ \text{mL}$ neutralise the acetic acid: $c(\text{CH}_3\text{COOH}) = \dfrac{0.100 \times 12.5}{25.0} = 0.0500\ \text{mol L}^{-1}$ .

(c) pH at the first end point. The acetic acid is now in $32.5\ \text{mL}$ : $c = 0.03846\ \text{mol L}^{-1}$ . Solving $\dfrac{x^2}{c - x} = K_a$ gives $[\text{H}^+] = 8.094 \times 10^{-4}\ \text{mol L}^{-1}$ , so $\text{pH} = 3.09$ . Sodium chloride does not change it.

(d) At the start. Only $\text{H}^+$ and $\text{Cl}^-$ count: $\kappa = \dfrac{0.0300 \times (349.6 + 76.3)}{1000}\ \text{S cm}^{-1} = 1.278\ \text{S m}^{-1}$ .

(e) At the second break. The volume is $45.0\ \text{mL}$ , with $\text{Na}^+$ $0.04444$ , $\text{Cl}^-$ $0.01667$ and $\text{CH}_3\text{COO}^-$ $0.02778\ \text{mol L}^{-1}$ . Then $1000\,\kappa = 2.2267 + 1.2717 + 1.1361 = 4.6344$ in S per cm, so $\kappa = 0.463\ \text{S m}^{-1}$ .

Check. A full model of the titration, solving the acetic acid equilibrium at every point of the burette, gives straight parts that meet at the same two end points, and the same two conductivities.

Watch out. In (c), using the acetic acid at its first strength, $0.0500\ \text{mol L}^{-1}$ , forgets the base already added and gives $\text{pH} = 3.03$ .

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