Question

6. A 0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910 M NaOH. If 14.20 mL of the NaOH solution is required to reach the second equivalence point, what is the molar mass of the acid?

Answer 1

Answer:

$MM_{acid}=96.9g/mol$

Explanation:

Hello,

In this case, it is more convenient to define this titration in terms of normality:

$eq-g_{acid}=eq-g_{NaOH}\\N_{acid}V_{acid}=N_{NaOH}V_{NaOH}$

In such a way, the normality of the acid, considering the normality of sodium hydroxide equals its molarity (one hydroxile in its structure) is:

$N_{acid}=\frac{N_{NaOH}V_{NaOH}}{V_{acid}}=\frac{0.1910N*14.20mL}{100.00mL} \\N_{acid}=0.0271\frac{eq-g}{L}$

Thus, the moles are:

$n_{acid}=0.0271\frac{eq-g}{L}*0.10000L*\frac{1mol}{2eq-g} =1.356x10^{-3}mol$

Hence, the molar mass:

$MM_{acid}=\frac{0.1310g}{1.356x10^{-3}mol} \\MM_{acid}=96.9g/mol$

Best regards.