OWLV2 Use the A student is asked to standardize a solution of calcium hydroxide.
ID: 581015 • Letter: O
Question
OWLV2 Use the A student is asked to standardize a solution of calcium hydroxide. He weighs out 1.02 g potassium hydrogen phthalate (KHC H 04, treat this as a monoprotic acid) It requires 31.0 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of the calcium hydroxide solution?M This calcium hydroxide solution is then used to titrate an unknown solution of perchloric acid B. If 24.8 mL of the calcium hydroxide solution is required to neutralize 20.7 mL of perchlorie acid, what is the molarity of the perchloric acid solution 8 more group attempts remainingExplanation / Answer
2KHC8H4O6 + Ca(OH)2 -----------> Ca(KC8H4O6)2 + 2H2O
2 moles 1 mole
no of moles of KHP = W/G.M.Wt
= 1.02/204.22 = 0.005 moles
2 moles of KHP react with 1 mole Ca(OH)2
0.005moles of KHP react with = 1*0.005/2 = 0.0025 moles
no of moles of Ca(OH)2 = molarity*volume in L
0.0025 = molarity*0.031
molarity = 0.0025/0.031 = 0.08 M
B. Ca(OH)2 + 2HClO4 -------------> Ca(ClO4)2 + 2H2O
1 mole 2 moles
Ca(OH)2 HClO4
M1 = 0.08M M2 =
V1 = 24.8ml V2 = 20.7ml
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
M2 = M1V1n2/V2
= 0.08*24.8*2/20.7 = 0.192M
2KHC8H4O6 + Ca(OH)2 -----------> Ca(KC8H4O6)2 + 2H2O
2 moles 1 mole
no of moles of KHP = W/G.M.Wt
= 0.981/204.22 = 0.0048 moles
2 moles of KHP react with 1 mole Ca(OH)2
0.0048moles of KHP react with = 1*0.0048/2 = 0.0024 moles
no of moles of Ca(OH)2 = molarity*volume in L
0.0024 = molarity*0.026
molarity = 0.0024/0.026 = 0.0923 M
B. Ca(OH)2 + 2HCl -------------> CaCl2 + 2H2O
1 mole 2 moles
Ca(OH)2 HCl
M1 = 0.0923M M2 =
V1 = 12.7ml V2 = 11.2ml
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
M2 = M1V1n2/V2
= 0.0923*12.7*2/11.2 = 0.21M
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