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

Next, we find the amount of NaOH that reacts with the acid H2X.

1. Calculate the moles of HCl used in neutralization:

   $n(HCl) = C \times V$

   • Where $C = 0.197 \, \text{M}$ and $V = 25.21 \, \text{mL} = 0.02521 \, \text{L}$

   Thus,

   $n(HCl) = 0.197 \times 0.02521 = 0.00497 \, \text{mol}$

2. Since the reaction ratio is 1:2 (HCl:NaOH), the moles of NaOH that reacted:

   $n(NaOH) = 2 \times n(HCl) = 2 \times 0.00497 = 0.00994 \, \text{mol}$

3. Therefore, the amount of NaOH that reacts with H2X is:

   $n(NaOH)_{reacting} = 0.0105 - 0.00497 = 0.00553 \, \text{mol}$

To find the molar mass of the acid H2X:

1. Using the formula,

   $M(H2X) = \frac{m}{n}$

   Where:

   • $m = 0.415 \, \text{g}$

   • $n = 2.765 \times 10^{-3} \, \text{mol}$ (which is obtained from earlier)

   2. Thus,

   $M(H2X) = \frac{0.415}{2.765 \times 10^{-3}} = 150.1 \, \text{g mol}^{-1}$