Question

6.74g of the monoprotic acid KHP(MW=204.1g/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

2KHP + Ca(OH)₂ → Ca(KP)₂ + 2H₂O

n(KHP) $= \frac{6.74}{204.1} = 0.033 \;mol$

n(Ca(OH)2) = $\frac{1}{2}$ n(KPH) $= \frac{1}{2} \times 0.033 = 0.016 \;mol$

Proportion:

0.703 mol – 1000 mL

0.016 mol – x mL

$x = \frac{0.016 \times 1000}{0.703} = 23.48 \;mL$

Answer: 23.48 mL