Question

The self ionisation constant of pure formic acid is 10−6M2 at room temperature. The density of formic acid is 1.22 g/cc. The percentage dissociation of formic acid x×10−y% then x+y is (where x,y are smaller whole numbers)

Answer 1

7

Answer 2

Solution Explanation:

1. Calculate the concentration of pure formic acid:
   Density = 1.22 g/cc = 1220 g/L
   Molar mass of HCOOH ≈ 46 g/mol

   $$c = \frac{1220}{46} \approx 26.52 \, \text{M}$$

2. Apply the self-ionisation equilibrium:
   For the dissociation, let α be the fraction of molecules ionised. Then:

   $$[H^+] = [HCOO^-] = c\alpha \quad \text{and} \quad K = (c\alpha)^2$$

   Given $K = 10^{-6}$ M²,

   $$c^2 \alpha^2 = 10^{-6} \quad \Rightarrow \quad \alpha^2 = \frac{10^{-6}}{(26.52)^2} \approx \frac{10^{-6}}{703.5} \approx 1.42 \times 10^{-9}$$ $$\alpha \approx \sqrt{1.42 \times 10^{-9}} \approx 3.77 \times 10^{-5}$$

3. Calculate percentage dissociation:

   $$\text{Percentage dissociation} = \alpha \times 100\% \approx 3.77 \times 10^{-3}\%$$

   Approximated in the format $x \times 10^{-y}\%$, we have $4 \times 10^{-3}\%$ (taking $x=4$ and $y=3$).

4. Find $x+y$:

   $$x+y = 4+3 = 7$$