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Question
Calculate the pH of the resultant mixtures: 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Solution
Moles of `"H"_3"O"^+` = `(2xx10xx0.1)/1000` = `.002 " mol"`
Moles of `"OH"^- = (10 xx 0.1)/1000 = 0.001 " mol"`
Excess of `"H"_2"O"^+` = .001 mol
Thus `["H"_3"O"^+] = .001/(20 xx 10^(-3)) = 10^(-3)/(20xx10^(-3))` = .05
`therefore "pH" = - log (0.05)`
= 1.30
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