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

a. 4,5x10⁻³ M

b. pH is 2,3

c. pOH is 11,7

Explanation:

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

0,53 M × = 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!

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