A student is asked to standardize a solution of sodium hydroxide. He weighs out
ID: 1052477 • Letter: A
Question
A student is asked to standardize a solution of sodium hydroxide. He weighs out 0.966 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 33.7 mL of sodium hydroxide to reach the endpoint. A. What is the molarity of the sodium hydroxide solution?
This sodium hydroxide solution is then used to titrate an unknown solution of perchloric acid. B. If 21.6 mL of the sodium hydroxide solution is required to neutralize 28.8 mL of perchloric acid, what is the molarity of the perchloric acid solution?
Explanation / Answer
A) moles of potassium hydrogen phthalate = 0.966 g / 204.22 g/mol
= 0.00473 mol
therefore
molarity of NaOH = moles of KHC8H4O4 / volume of KHC8H4O4
= 0.00473 mol / ( 33.7 x 10^-3 L )
= 0.14 mol / L
= 0.14 M
B) rom molarity equation , we have
M1 V1 = M2 V2
M2 = M1 * V1 / V2
= 0.14 M x 21.6 ml / 28.8 ml
= 0.105 M
therefore
the molarity of the perchloric acid solution is 0.105 M.
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