A student is asked to standardize a solution of sodium hydroxide. He weighs out
ID: 1037172 • Letter: A
Question
A student is asked to standardize a solution of sodium hydroxide. He weighs out 0.920 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 29.5 mL of sodium hydroxide to reach the endpoint.
A.) What is the molarity of the sodium hydroxide solution? 0.153 M This sodium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.
B.) If 10.7 mL of the sodium hydroxide solution is required to neutralize 25.0 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution? 0.357 M
Explanation / Answer
KHC8H4O4 + NaOH -----------------> NaKC8H4O4 + H2O
no of moles of KHC8H4O4 = W/G.M.Wt
= 0.92/204.22 = 0.0045 moles
1 moles of KHC8H4O4 react with 1 mole of NaOH
0.0045 moles of KHC8H4O4 react with 0.0045 moles of NaOH
no of moles of NaOH = molarity * volume in L
0.0045 = molarity *0.0295
molarity = 0.0045/0.0295 = 0.153M
molarity of NaOH = 0.153M
B. NaOH + HI --------------------> NaI + H2O
NaOH HI
M1 = 0.153M M2 =
V1 = 10.7ml V2 = 25ml
n1 = 1 n2 =1
M1V1/n1 = M2V2/n2
M2 = M1V1n2/V2n1
= 0.153*10.7*1/25*1 = 0.0654M
molarity of HI = 0.0654M
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