Using Titration Analysis to Determine pH Change

Titration Analysis: Calculation of pH Change

When a 15.6 mL sample of a 0.334M aqueous hypochlorous acid solution is titrated with a 0.474M aqueous sodium hydroxide solution, what is the pH after 16.5 mL of sodium hydroxide have been added?

Final answer:

The pH after 16.5 mL of sodium hydroxide have been added to a 15.6 mL sample of a 0.334M aqueous hypochlorous acid solution is 4.46.

Explanation:

To determine the pH after 16.5 mL of sodium hydroxide have been added to a 15.6 mL sample of a 0.334M aqueous hypochlorous acid solution, we need to consider the reaction between hypochlorous acid (HOCl) and sodium hydroxide (NaOH). The balanced equation for the reaction is:

HOCl(aq) + NaOH(aq) -> H2O(l) + NaOCl(aq)

Hypochlorous acid is a weak acid, so we need to consider the dissociation of HOCl in water. The dissociation constant (Ka) for HOCl is 3.5 * 10^-8. To determine the pH after the titration, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base (NaOCl) and [HA] is the concentration of the weak acid (HOCl).

From the information given, we can calculate the moles of NaOH added:

0.474 mol/L * 0.0165 L = 0.0078 mol

The initial moles of HOCl in the sample:

0.334 mol/L * 0.0156 L = 0.00521 mol

Since NaOH reacts with HOCl in a 1:1 ratio, the number of moles of NaOCl formed is also 0.0078 mol.

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA]) = -log(Ka) + log([NaOCl]/[HOCl])

To find the concentrations of NaOCl and HOCl, divide the moles by the total volume of the solution (15.6 mL + 16.5 mL = 32.1 mL = 0.0321 L):

[NaOCl] = 0.0078 mol / 0.0321 L = 0.243 M

[HOCl] = 0.00521 mol / 0.0321 L = 0.162 M

Plugging these values into the Henderson-Hasselbalch equation:

pH = -log(3.5 * 10^-8) + log(0.243/0.162) = 4.46

Therefore, the pH after 16.5 mL of sodium hydroxide have been added is 4.46.