Answer:
[tex]V=0.0310L=3.10mL[/tex]
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:
[tex]n_{acid}=n_{KOH}[/tex]
Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:
[tex]n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol[/tex]
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:
[tex]V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL[/tex]
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