A student is asked to standardize a solution of calcium
hydroxide. He weighs out 0.921g potassium
hydrogen phthalate (KHC8H4O4,
treat this as a monoprotic acid).
It requires 17.3 mL of calcium
hydroxide to reach the endpoint.
A. What is the molarity of thecalcium
hydroxide solution?
M=?
his calcium hydroxidesolution is then used to
titrate an unknown solution ofhydrobromic
acid.
B. If 19.5 mL of the
calcium hydroxide solution is required to
neutralize 14.2 mL ofhydrobromic
acid, what is the molarity of the hydrobromic
acid solution?
M=?
a)
Molarity of [Ca(OH)2]
mol of acid = mass/MW = 0.921/204.22 = 0.00451 mol of acid
ratio is:
2 mol of acid = 1 mol of base
0.00450 --> 0.00450/2 = 0.00225 mol of base
V = 17.3 mL
[Ca(OH)2] = mol /V = (0.00225) /(17.3*10^-3) = 0.13 M
B)
V = 19.5 mL of base is used, and
V = 14.2 mL of HBr required...
ratio is
1 mol of base = 2 mol of acid
mol of base = MV = 0.13*19.5 = 2.535*10^-3 mol of base
mol of acid = 2x mol of base = 2*2.535*10^-3 = 0.00507 mol of acid
V acid = 14.2 mL
[HBr] = mol/V =(0.00507)/(14.2*10^-3) = 0.3570 M of HBr
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