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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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.