Question

A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid if 21.5 mL of the NaOH solution is required to neutralize the sample?

Answer 1

Answer:

$MM_{acid}=140.1g/mol$

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

$n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}$

Thus, solving for the moles of the acid, we obtain:

$n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol$

Then, by using the mass of the acid, we compute its molar mass:

$MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol$

Regards.