Question

For the titration of 25.00 mL of 0.1000 M HCl with 0.1000 M NaOH, calculate the pH of the reaction mixture after each of the following total volumes of base have been added to the original solution. (Remember to take into account the change in total volume.) Select a graph showing the titration curve for this experiment. (a) 0 mL (b) 10.00 mL (c) 24.90 mL (d) 24.99 mL (e) 25.00 mL (f) 25.01 mL (g) 25.10 mL (h) 26.00 mL (i) 50.00 mL

Answer 1

The acid base reaction is given as

HCl (aq) + NaOH (aq) -------> NaCl (aq) + H₂O

As per the stoichiometric equation,

1 mole HCl = 1 mole NaOH.

(a) No NaOH is added; the solution contains only HCl. The pH of the solution is given as

pH = -log [H⁺] = -log (0.1000) = 1.00 (ans) (remember, for a strong acid like HCl, [HCl] =[H⁺])

(b) Millimoles of HCl taken = (25.00 mL)*(0.1000 M)*(1 mol/L/1 M) = 2.5 mmole.

Millimoles of NaOH added = millimoles HCl neutralized = (10.00 mL)*(0.1000 M)*(1 mol/L/1 M) = 1.0 mmole.

Millimoles HCl retained at equilibrium = (2.5 – 1.0) mmole = 1.5 mmole.

Total volume of solution = (25.00 + 10.00) mL = 35.00 mL.

[HCl] = [H⁺] = (1.5 mmole)/(35.00 mL) = 0.04286 M and pH = -log [H⁺] = -log (0.04286) = 1.3679 ≈ 1.37 (ans).

(c) Millimoles of NaOH added = millimoles HCl neutralized = (24.90 mL)*(0.1000 M)*(1 mol/L/1 M) = 2.49 mmole.

Millimoles HCl retained at equilibrium = (2.5 – 2.49) mmole = 0.01 mmole.

Total volume of solution = (25.00 + 24.90) mL = 49.90 mL.

[HCl] = [H⁺] = (0.01 mmole)/(49.90 mL) = 2.004*10^-4 M and pH = -log [H⁺] = -log (2.004*10^-4) = 3.6981 ≈ 3.70 (ans).

(d) Millimoles of NaOH added = millimoles HCl neutralized = (24.99 mL)*(0.1000 M)*(1 mol/L/1 M) = 2.499 mmole.

Millimoles HCl retained at equilibrium = (2.5 – 2.499) mmole = 0.001 mmole.

Total volume of solution = (25.00 + 24.99) mL = 49.99 mL.

[HCl] = [H⁺] = (0.001 mmole)/(49.99 mL) = 2.0004*10^-5 M and pH = -log [H⁺] = -log (2.0004*10^-5) = 4.6989 ≈ 4.70 (ans).

(e) Millimoles of NaOH added = millimoles HCl neutralized = (25.00 mL)*(0.1000 M)*(1 mol/L/1 M) = 2.5 mmole.

HCl is completely neutralized and the solution contains NaCl and H₂O. This is a neutral solution and the pH is 7.00 (ans).

(f) Volume of NaOH added beyond the equivalence point = (25.01 – 25.00) mL = 0.01 mL.

Millimoles of NaOH added in excess = (0.01 mL)*(0.1000 M)*(1 mol/L/1 M) = 0.001 mmole.

Total volume of solution = (25.00 + 25.01) mL = 50.01 mL.

The solution now contains NaOH and is a basic solution; we shall find out the pOH therefore.

We have [NaOH] = [OH^-] = (0.001 mmole)/(50.01 mL) = 1.9996*10^-5 M and pOH = -log [OH^-] = -log (1.9996*10^-5) = 4.6690.

We know that pH + pOH = 14.00; therefore, pH = 14.00 – pOH = 14.00 – 4.6690 = 9.3310 ≈ 9.33 (ans).

(g) Volume of NaOH added beyond the equivalence point = (25.10 – 25.00) mL = 0.10 mL.

Millimoles of NaOH added in excess = (0.10 mL)*(0.1000 M)*(1 mol/L/1 M) = 0.01 mmole.

Total volume of solution = (25.00 + 25.10) mL = 50.10 mL.

The solution now contains NaOH and is a basic solution; we shall find out the pOH therefore.

We have [NaOH] = [OH^-] = (0.01 mmole)/(50.10 mL) = 1.996*10^-4 M and pOH = -log [OH^-] = -log (1.996*10^-4) = 3.6998.

We know that pH + pOH = 14.00; therefore, pH = 14.00 – pOH = 14.00 – 3.6998 = 10.3002 ≈ 10.30 (ans).

(h) Volume of NaOH added beyond the equivalence point = (26.00 – 25.00) mL = 1.00 mL.

Millimoles of NaOH added in excess = (1.00 mL)*(0.1000 M)*(1 mol/L/1 M) = 0.1 mmole.

Total volume of solution = (25.00 + 26.00) mL = 51.00 mL.

The solution now contains NaOH and is a basic solution; we shall find out the pOH therefore.

We have [NaOH] = [OH^-] = (0.1 mmole)/(51.00 mL) = 0.001961 M and pOH = -log [OH^-] = -log (0.001961) = 2.7075.

We know that pH + pOH = 14.00; therefore, pH = 14.00 – pOH = 14.00 – 2.7075 = 11.2925 ≈ 11.29 (ans).

(i) Volume of NaOH added beyond the equivalence point = (50.00 – 25.00) mL = 25.00 mL.

Millimoles of NaOH added in excess = (25.00 mL)*(0.1000 M)*(1 mol/L/1 M) = 2.5 mmole.

Total volume of solution = (25.00 + 50.00) mL = 75.00 mL.

The solution now contains NaOH and is a basic solution; we shall find out the pOH therefore.

We have [NaOH] = [OH^-] = (2.5 mmole)/(75.00 mL) = 0.0333 M and pOH = -log [OH^-] = -log (0.0333) = 1.4775.

We know that pH + pOH = 14.00; therefore, pH = 14.00 – pOH = 14.00 – 1.4775 = 12.5225 ≈ 12.52 (ans).

Tabulate the pH vs volume of NaOH added.

Volume of NaOH added (mL)	pH
0.00	1.00
10.00	1.37
24.90	3.70
24.99	4.70
25.00	7.00
25.01	9.33
25.10	10.30
26.00	11.29
50.00	12.52

Plot of pH vs volume of NaOH for titration of HCl with NaOH