Question

Given the following information: 1.6 g of an unknown monoprotic acid (HA) required 50.80 mL of a 0.35 M NaOH solution to reach the equivalence point, calculate the molar mass (g/mol) of the acid. Enter the value ONLY. Do not include the units.

Answer 1

Answer: 89.9

Explanation:-

The balanced chemical equation is :

$NaOH(aq)+HA(aq)\rightarrow NaAaq)+H_2O(l)$

To calculate the moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$     .....(1)

Molarity of $NaOH$ solution = 0.35 M

Volume of solution = 50.80 ml

Putting values in equation 1, we get:

$0.35M=\frac{\text {moles of} NaOH\times 1000}{50.80ml}$

${\text {moles of} NaOH=0.0178$

According to stoichiometry:

1 mole of NaOH reacts with = 1 mole of HCl

Thus 0.0178 moles of NaOH reacts with = $\frac{1}{1}\times 0.0178=0.0178$ moles of HCl

mass of HCl = $moles\times {\text {Molar mass}}$

1.6g=  $0.0178\times {\text {Molar mass}}$

Molar mass of HCl = 89.9

The molar mass (g/mol) of the acid is 89.9