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A beaker containing HCl solution is neutralized with 0.7 M KOH solution. What do you know about the ratio [H₃O]⁺/[OH]^-in the reaction mixture at the end point?

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The same HCl solution can also be neutralized with 0.7 M Ca(OH)₂ solution. What would be the ratio [H₃O]⁺/[OH]^- in the reaction mixture at the end point in this case?

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Answer 1

1) HCl + KOH -> H2O + KCl

neutralized means pH = 7 this means [H3O+] = [OH-]

[H3O+]/[OH-] = 1

2)if you use the same amount of HCl as in 1) you get:

HCl + Ca(OH)2 -> Ca²⁺ + Cl^- + H2O+ OH-

that means we have more OH- than H3O+ in the solution and the ratio is >1, the pH will be over 7