Answered

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 :

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

[tex]2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2[/tex]

Moles of KHP = [tex]\frac{6.74 g}{204.2 g/mol}=0.0330 mol[/tex]

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

[tex]\frac{1}{2}\times 0.0330 mol=0.01650 mol[/tex] of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]

[tex]0.703 M=\frac{0.01650 M}{V}[/tex]

[tex]V=\frac{0.01650 M}{0.703 M}=0.0235 mL[/tex]

Volume of the calcium hydroxide solution used is 0.0235 mL.

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