Question

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of the acid to the equivalence point.

Answer 1

Answer:

$V=0.0310L=3.10mL$

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$

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