Question

# A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.925 g...

A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.925 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 28.3 mL of calcium hydroxide to reach the endpoint.

A. What is the molarity of the calcium hydroxide solution? This calcium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.

B. If 13.3 mL of the calcium hydroxide solution is required to neutralize 29.9 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution?

KHC8H4O4 has only one acidic proton

2HA + Ca[OH]2 ------> CaA2 + 2H2O

2 mole of HA can be neutralised by 1 mole of Ca[OH]2

MOles of KHC8H4O4 = mass/ molar mass = 0.00453

0.00453 moles of HA is neutralised by ==> 0.00453/2 = 0.002265 moles of Ca[OH]2

A] Moles = Molarity*V in L

M = 0.002265 / 28.3 = 0.08 M

B] 2HI + Ca [OH]2 -----> CaI2 + 2H2O

2 moles of HI neutralised by 1 mole of Ca[OH]2

13.3*0.08 = 1.064 millimoles of Ca[OH]2 can neutralize ==== > 1.064*2 = 2.128 milli moles of HI

MV = 2.128

M = 2.128 / 29.9 = 0.07117 M

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