A solution of 0.357 M HClO (Ka = 3.5 x 10-8) with volume 75. mL is mixed with 50. mL of a 0.800 M KOH solution. What is the final pH?
Moles of HClO = 0.357 M x 75. mL = 27 mmol
Moles of KOH = 0.800 M x 50. mL = 40 mmol
HClO + KOH KClO + H2O
Theoretically, 1 mol of HClO gets completely neutralized by 1 mol of KOH.
Therefore, 27 mmol of HClO is completely neutralized by 27 mmol of KOH.
Hence, the unreacted KOH in the solution = (40 - 27) = 13 mmol
Total volume of the solution after mixing = (75. + 50.) mL =125 mL
Concentration of unreacted KOH = 13 mmol/125 mL
= 0.10 M
KOH is a strong base. Thus, 0.10 M of KOH completely dissociate in water to give 0.10 M of OH- ions.
Now,
pOH = - log[OH-]
= - log(0.10)
= 1.0
And,
pH = 14 - pOH
= 14 - 1.0
= 13
Hence, the final pH of the solution = 13
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