Question

What is the pH (to nearest 0.01 pH unit) of a solution prepared by mixing 80.0 mL of 0.400 M HCl and 54.0 mL of 0.673 M NaOH?

Answer 1

Consider the neutralization reaction.

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

As per the balanced stoichiometric equation,

1 mole HCl = 1 mole NaOH.

Millimoles of HCl taken = (volume of HCl in mL)*(concentration of HCl in mol/L) = (80.0 mL)*(0.400 M)*(1 mol/L/1 M) = 32.0 mmole (1 M = 1 mol/L).

Millimoles of NaOH taken = (54.0 mL)*(0.673 M)*(1 mol/L/1 M) = 36.342 mmole.

We note that the millimoles of NaOH exceeds the millimoles of HCl. Since NaOH and HCl react on a 1:1 molar ratio, hence, we can say that NaOH has completely neutralized the HCl and the solution contains excess NaOH.

Millimoles of excess NaOH = (36.342 – 32.0) mmole = 4.342 mmole.

Total volume of the solution = (80.0 + 54.0) mL = 134.0 mL.

Concentration of excess NaOH = (millimoles of excess NaOH)/(total volume of solution) = (4.342 mmole)/(134.0 mL) = 0.0324 mmol/mL ≈ 0.0324 M.

Since we have excess NaOH, we must find the pOH of the solution. The pOH is defined as pOH = -log [OH^-] = -log (0.0324) = 1.4894.

We know that pH + pOH = 14; therefore, pH = 14 – pOH = 12.5106 ≈ 12.51 (ans).