Question

# A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.988 g...

A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.988 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 18.7 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 12.7 mL of the calcium hydroxide solution is required to neutralize 14.9 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution?

the reaction between KHC8H4O4 and Ca(OH)2 is

2 KC8H5O4 + Ca(OH)2 >> Ca( KC8H4O4 )2 + 2 H2O

2 moles of KHP requires 1 mole of Ca(OH)2 , molar mass of KHP= 204.22 g/mole, moles =mass/molar mass

moles of KHP= 0.988/204.22 =0.004838, moles of Ca(OH)2= 0.004838/2= 0.002419

volume of Ca(OH)2 consumed= 18.7ml, 1000ml =1 L, 18.7 ml= 18.7/1000L=0.0187 L

Molarity of Ca(OH)2= moles of Ca(OH)2/ volume in L= 0.002419/0.0187= 0.1293 M

the reaction between Ca(OH)2 and HCl is Ca(OH)2+2HCl------>CaCl2+ 2H2O

1 moles of Ca(OH)2 requires 2 moles of HCl, moles of Ca(OH)2= molarity* volume in L= 0.1293*12.7/1000

moles of HCl = 0.1293*12.7/(2000)

molarity of HCl = moles/Volume in L= (0.1293*12.7/2000)/(14.9/1000)=0.055M

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