Question

You add 1.00 mL of 0.400 M NaOH to 50.00 mL of pure water, and to this mixture you then add 11.00 mL of 0.600 M HCl. What will be the pH of the resulting solution?

Answer 1

When 1.00 mL of 0.400 M NaOH is added to 50.00 mL of pure water, the solution becomes alkaline. Next, 11.00 mL of 0.600 M HCl is added to the solution. The alkaline solution is neutralized (partly or completely) by the acid. The extent of neutralization is guided by the number of mols (or millimols) of the base (NaOH) and acid (HCl) present in the solution.

Millimols NaOH = (1.00 mL)*(0.400 M) = 0.400 mmol.

Millimols HCl = (11.00 mL)*(0.600 M) = 6.600 mmol.

NaOH reacts with HCl as per the equation below.

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

As per the stoichiometric equation,

1 mol NaOH = 1 mol HCl.

Therefore, 0.400 mmol NaOH = 0.400 mmol HCl.

NaOH is completely neutralized by HCl and the amount of left over HCl in the solution = (6.600 – 0.400) mmol = 6.200 mmol.

Total volume of the solution = (1.00 + 50.00 + 11.00) mL = 62.00 mL.

Molarity of left over HCl in the solution = (millimols HCl)/(volume of the solution)

= (6.200 mmol)/(62.00 mL)

= 0.100 M.

The pH of the solution is given as

pH = -log [H+]

= -log (0.100 M)

= 1.00 (HCl is completely ionized in aqueous solution).

The pH of the solution = 1.00 (ans).