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A 3.458 g sample of KHP, a monoprotic acid, requires 45.71 mL of a KOH solution to reach the endpoint. What is the concentration of the KOH solution? The molar mass of KHP is 204.22 g/mol.

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Answer:

[tex]M_{KOH}=0.3704M[/tex]

Explanation:

Hello there!

In this case, since the titration of bases when using monoprotic acids like KHP, occurs in a 1:1 mole ratio, it is possible to use the following equation, because at the endpoint the moles of the KHP and KOH get equal:

[tex]n_{KHP}=n_{KOH}[/tex]

In such a way, we first calculate the moles of KOH given the mass and molar mass of KHP:

[tex]n_{KHP}=n_{KOH}=3.458g*\frac{1mol}{204.22g}=0.0169mol[/tex]

Next, since we have the volume of KOH, we first take it to liters (0.04571 L) to that we obtain the following concentration:

[tex]M_{KOH}=\frac{0.0169mol}{0.04571L}\\\\M_{KOH}=0.3704M[/tex]

Regards!

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