I mix together 50.0 mL of 0.100 M NaIO3, 50.00 mL of 0.100 M NaOH, and 10.0 mL of 0.100 M HIO3. What is the pH of the mixture

pH = 12.66

Explanation:

The HIO3 reacts with NaOH as follows:

HIO3 + NaOH → H2O + NaIO3

The moles of HIO3 and NaOH added are:

Moles HIO3:

0.0100L * (0.100mol / L) = 0.00100 moles HIO3

Moles NaOH:

0.05000L * (0.100mol / L) = 0.00500 moles NaOH

As moles NaOH > Moles HIO3, the moles of NaOH that remain are:

0.00500mol - 0.00100mol = 0.00400 moles NaOH.

After the reaction you will have only NaOH and NaIO3. As NaIO3 is a salt, the pH of the solution is determined by only NaOH. Its concentration is:

Moles NaOH: 0.00500 moles NaOH

Volume: 50.0mL + 50.0mL + 10.0mL = 110.0mL = 0.110L

Molarity: 0.0455M NaOH = [OH-]

pOH = -log [OH-] = 1.34

pH = 14 - pOH