Question 8 0.415 g of a pure acid, H₂X(s), is added to exactly 100 mL of 0.105 M NaOH(aq) - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

First, calculate the moles of NaOH in excess using the HCl neutralization:

$n( ext{NaOH}) ext{ in excess} = C imes V = 0.197 imes 0.02521 = 0.00497 ext{ mol}$

Now, we can find the amount of NaOH that reacts with H₂X using:

$n( ext{NaOH}) ext{ reacting} = n( ext{NaOH}) ext{ added} - n( ext{NaOH}) ext{ in excess}$

So:

$n( ext{NaOH}) ext{ reacting} = 0.0105 - 0.00497 = 0.00553 ext{ mol}$

Using the number of moles calculated:

The molar mass of H₂X can be calculated by rearranging the molar mass formula:

$M(H₂X) = \frac{m}{n}$

Where:

• m is the mass of the acid (0.415 g)
• n is the moles of H₂X which is half of the moles of NaOH reacting:

$n(H₂X) = \frac{1}{2} n( ext{NaOH}) = \frac{1}{2} imes 0.00553 = 0.002765$

Thus:

$M(H₂X) = \frac{0.415}{0.002765} = 150 ext{ g mol}^{-1}$