Calculating the Ph of a solution?

Question

Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M NaOH and 25.0 mL of 0.10 M HClO. (Ka= 3.5 × 10-8 for HClO.)

a. ) 1.18

b. ) 1.48

c. ) 11.73

d. ) 12.52

e. ) 12.82

Answer 1

(0.050 L) (0.10 mol/L) = 0.0050 moles of -OH added

(0.025 L) (0.10 mol/L) = 0.0025 moles of acid added

Since 2x times as much base is added than acid, all the acid has been neutralized, and so it doesn't matter what the pKa of the acid is. After the reaction, there are  0.0050 - 0.0025 = 0.0025 moles of -OH remaining, so

[-OH] = (0.0025 moles) / (0.050 L + 0.025 L) = 0.033 M

pOH = -log[-OH] = 1.48; so pH = 12.52. Answer is d.).