A scientist dilutes 50.0 mL of a pH 5.85 solution of HCl to 1.00 L. What is the pH of the diluted solution (Kw = 1.0 × 10⁻¹⁴)?

Question

A scientist dilutes 50.0 mL of a pH 5.85 solution of HCl to 1.00 L. What is the pH of the diluted solution (Kw = 1.0 × 10⁻¹⁴)?

It says I also have to add the H^+ from the water into my calculation im not sure how to do that.

Answer 1

From the pH = 5.85, we can find the moles of H+ present from the HCl.

Since pH = -log [H⁺], then ...

[H⁺] = 1x10^-5.85 = 1.41x10^-6 M

Diluting this solution (50 ml to 1000 ml) we have ...

(50 ml)(1.41x10^-6 M) = (1000 ml)(x M)

x = 7.05x10^-8 M H⁺ (this is the H⁺ concentration contributed by the HCl)

Because this is such a low value of H⁺, we need to also include the H⁺ that comes from the auto-ionization of water.

H₂O ==> H⁺ + OH^-

Kw = 1x10^-14

[H⁺] = [OH^-] = 1x10^-7 M

7.05x10^-8 M + 1.00x10^-7 M = 2.95x10^-8 M [H⁺]

pH = -log 2.95x10^-8

pH = 7.53