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16 days ago

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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?

1 answer:

eduard [2.5K]16 days ago

4 0

Answer:

The volume of calcium hydroxide solution utilized is 0.0235 mL.

Explanation:

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

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

In accordance with the reaction, 2 moles of KHP react with 1 mole of calcium hydroxide, thus 0.0330 moles of KHP will react with;

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

The molarity of calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

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

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

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

The volume of the calcium hydroxide solution utilized is 0.0235 mL.

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