Question

(b) What is the pH of 8× 10–8 M HCl?

answer is not

7.09 or 6.96

"Incorrect. Consider the H contributed by both the acid and the water."

Answer 1

We know that pure water undergoes self-ionization to form Hydronium ion and hydroxide ion as

2H₂O_(l) <=> H₃O⁺_(aq) + OH^-_(aq)

So for this reaction, Kw = 10^-14

Kw = [H₃O⁺][OH^-] = 10^-14

So [H₃O⁺] = (10^-14)^1/2 = 10^-7

Therefore, the pH of pure water = -log([H₃O⁺]) = -log(10^-7) = 7

We know that HCl is strong acid which dissociated completely in the water to form the hydronium ion and chloride ion. The reaction is as

HCl_(aq) + H₂O_(l) --> H₃O⁺_(aq) + Cl^-_(aq)

From the above reaction,

[H₃O⁺] = [HCl] = 8.0 *10^-8 M

As we know that water will still self-ionize to produce equal concentration of hydronium ion and hydroxide ions. If in the acidic solution, the self-ionization of water generates xM of hydronium ions and xM of hydroxide ions, then

(x + 8.0 *10^-8) * x = 10^-14

Where (x + 8.0 x*10^-8) represents the equilibrium concentration of the hydronium ion and x represents the equilibrium concentration of hydroxide ions.

The above equation can also be written as

[(x + 8.0 *10^-8) * x] - 10^-14 = 0 ; or

x² + (8.0 *10^-8)x - 10^-14 = 0

Solving above equation, we get-

x = 6.7703*10^-8 M

Therefore, the equilibrium concentration of hydronium ions will be

[H₃O⁺] = (8.0 *10^-8) + (6.7703*10^-8) = 1.477*10^-7

Therefore, the pH of the solution will be

pH = -log([H₃O⁺])

= -log(1.477*10^-7)

= 6.83

Hence the pH of 8.0 *10^-8 M HCl solution = 6.83