Question

8.5 mL of a 0.53 M solution of the strong acid HCl was diluted with water to a total volume of 1000.0 mL at 30°C. Calculate the following. (The Kw at 30°C is 1.46 ✕ 10−14.)

(a) the concentration of H3O+ ions in the diluted HCl solution M
(b) the pH of the diluted solution (c) the pOH of the diluted solution

Answer 1

a. 4,5x10⁻³ M

b. pH is 2,3

c. pOH is 11,7

Step-by-step explanation:

The dilution of the 0,53 M solution of the strong acid gives a concentration of:

0,53 M ×
`(8,5mL)/(1000,0mL)` = 4,5x10⁻³ M of HCl

a. The H⁺ ions from HCl reacts with water as the following:

H⁺ + H₂O → H₃O⁺

Thus, molar concentration of H₃O⁺ in the diluted solution is 4,5x10⁻³ M

b. The pH is defined as -log [H₃O⁺], thus, pH of this solution is:

-log [4,5x10⁻³ M] = 2,3

c. As pH + pOH = 14, pOH is:

14- pH = 14 - 2,3 = 11,7

I hope it helps!