An aqueous solution contains 0.352 M hydrocyanic acid. Calculate the pH of the solution after the...
An aqueous solution contains 0.352 M
hydrocyanic acid.
Calculate the pH of the solution after the addition of
4.45E-2 moles of potassium
hydroxide to 250 mL of this
solution.
(Assume that the volume change does not change upon adding
potassium hydroxide)
Homework Answers
HCN + KOH ---------------> KCN + H2O
[HCN] = 0.352M,
[KOH] = 4.45 x 10-2 x 1000/250 = 17.8 x 10-2 = 0.178M
Acid concentration > base concentration
[H+]= 0.352 - 0.178 /2 = 0.174/2M = 0.087M
PH = -log[0.087] = 1.06
PH = 1.06
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