Question

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of calcium hydroxide to the equivalence point. What volume of base was used?

Answer 1

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

Volume of the calcium hydroxide solution used is 0.0235 mL.