Question

The hydrogen ion concentration of a $ 10^{-8}M HCl $ aqueous solution at 298 K $ ( K_w = 10^{-14} ) $ is :-

• A. $ 1.0 \times 10^{-6} M $
• B. $ 1.0525 \times 10^{-7} M $
• C. $ 9.525 \times 10^{-8} M $
• D. $ 1.0 \times 10^{-8} M $

Answer 1

The Correct Option is B

To calculate the hydrogen ion concentration in a diluted acid solution like this, we need to consider both the contribution from the acid and the autoionization of water.

1. Given: The concentration of HCl is 10^{-8} \, M.
2. HCl is a strong acid, meaning it ionizes completely in water. Thus, we initially assume that the concentration of hydrogen ions [\text{H}^+] from HCl is 10^{-8} \, M.
3. However, since this is a very dilute solution (close to the concentration of water's own ionization), we must also consider the contribution of [\text{H}^+] from the autoionization of water. The ionic product of water at 298 K is given by K_w = 10^{-14}.
4. The equation for the autoionization of water is:
   K_w = [\text{H}^+][\text{OH}^-]
   Since the contribution from autoionization is significant here, we represent the total [\text{H}^+] concentration as
   [\text{H}^+] = x + 10^{-8}
   , where x is the concentration of hydrogen ions from the autoionization of water.
5. From the autoionization of water, we know
   [\text{H}^+][\text{OH}^-] = 10^{-14}
   Setting [\text{H}^+] from autoionization equal to [\text{OH}^-] gives x^2 = 10^{-14}, hence x = 10^{-7} \, M
6. Total [\text{H}^+] is
   [\text{H}^+] = 10^{-8} + 10^{-7}
   which simplifies to 1.0525 \times 10^{-7} \, M when accounting for appropriate significant figures.
7. Thus, the correct answer is 1.0525 \times 10^{-7} \, M.

This value accounts for both the contribution from the dissociated strong acid and the equilibrium condition of water, which is significant in dilute solutions like this.