The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is .

7

Explanation:

This question is a neutralization reaction.

50 ml of 0.125 M of KOH = (50 x 0.125) = 6.25 ml

In this same way, 50 ml of 0.125M of HCl = 50x0.125

= 6.25 ml

Then KOH is going to neutralise the HCl fully suvh that the pH of the medium is going to be 7

(H+) + (OH-) = H2O

Then Kw = [H=][OH-]

= [H+]² = 10^-14

Such that [H+] = 10^-7

The pH would be = -log [H+]

= -log[10^-7]

When inputted on a calculator this gives us 7

= 7

This is our answer

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