A student is asked to standardize a solution of sodium hydroxide . He weighs out
ID: 474882 • Letter: A
Question
A student is asked to standardize a solution of sodium hydroxide. He weighs out 0.987 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 29.9 mL of sodium hydroxide to reach the endpoint.
A. What is the molarity of the sodium hydroxide solution? M
This sodium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.
B. If 19.3 mL of the sodium hydroxide solution is required to neutralize 13.4 mL of hydroiodic acid, what is the molarity of the hydroiodic acidsolution? M
Explanation / Answer
Answer(a)
2KHP + Ca(OH)2 ==> 2H2O + CaK2P2
0.987 g KHP x (1 mole KHP / 204.2 g KHP) =
0.00483 moles KHP
0.00483 moles KHP x (1 mole Ca(OH)2 / 2 moles KHP) = 0.00240 moles Ca(OH)2
moles Ca(OH)2 = M Ca(OH)2 x L Ca(OH)2
0.00240 = (M Ca(OH)2)(0.0299 L )
0.0803 = M Ca(OH)2
Answer(b)
2HI + Ca(OH)2 ==> 2H2O + CaI2
moles Ca(OH)2 = M Ca(OH)2 x L Ca(OH)2
= (0.0803)(0.0193) = 0.0015 moles Ca(OH)2
0.0015 moles Ca(OH)2 x (2 moles HI / 1 mole Ca(OH)2) = 0.0031 moles HI
moles HI = M HI x L HI
0.0031= (M HI)(0.0134 L)
M HI = 0.2313
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