1. A student is asked to standardize a solution of sodium hydroxide . He weighs
ID: 1047756 • Letter: 1
Question
1. A student is asked to standardize a solution of sodium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 20.5 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 hydrobromic acid.
B. If 20.5 mL of the sodium hydroxide solution is required to neutralize 10.6 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution?
2. A student is asked to standardize a solution of barium hydroxide. He weighs out 0.946 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 25.7 mL of barium hydroxide to reach the endpoint.
A. What is the molarity of the barium hydroxide solution?
This barium hydroxide solution is then used to titrate an unknown solution of hydrobromic acid.
B. If 19.5 mL of the barium hydroxide solution is required to neutralize 18.6 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution?
Explanation / Answer
A.
moles of KHC8H4O4 = 1.08 / 204.22 = 5.29 x 10^-3
KHC8H4O4 (aq) + NaOH (aq) -------------> NaKC8H4O4 (aq) + H2O
according to balanced equation:
moles of KHC8H4O4 = moles of NaOH
5.29 x 10^-3 = M x 20.5 x 10^-3
M = 0.258
molarity of NaOH = 0.258 M
B)
NaOH + HBr -------------> NaBr + H2O
moles of NaOH = 20.5 x 0.258 /1000 = 5.29 x 10^-3
moles of NaOH = moles of HBr
5.29 x 10^-3 = M x 10.6 x 10^-3
M = 0.499 M
molarity of HBr = 0.499 M
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