Question

. A 0.264 g sample of an unknown monoprotic acid (HA) is neutralized by 38.2 mL of 0.0800 M NaOH. Calculate the molar mass of the acid. HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)

Answer 1

Answer:- 86.4 gram per mol.

Solution:- From given balanced equation, the acid and NaOH reacts in 1:1 mol ratio. So, moles of acid are exactly equal to the moles of NaOH used for the titration.

Moles of NaOH could be calculated from it's given volume and molarity.

$38.2mL(\frac{1L}{1000mL})(\frac{0.0800molNaOH}{1L})(\frac{1molHA}{1molNaOH})$

= 0.003056 mol HA

There are 0.264 g of the sample of HA titrated with NaOH and we know that molar mass unit is gram per mol. So, let's divide the grams by the calculated moles of the acid.

molar mass of acid,HA = $\frac{0.264g}{0.003056mol}$

= $\frac{86.4g}{mol}$

Hence, the molar mass of the acid, HA is 86.4 gram per mol.