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Question
Calculate the pH of the resultant mixtures: 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl.
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Solution
Moles of `"H"_3"O"^+ = (25 xx 0.1)/1000` = .0025 mol
Moles of `"OH"^(-) = (10 xx 0.2 xx 2)/1000 = .0040 " mol"`
Thus, excess of `"OH"^-` = .0015 mol
`["OH"^(-)] = .0015/(35 xx 10^(-3)) "mol/L" = .0428`
`"pOH" = - log ["OH"]`
= 1.36
pH = 14 - 1.36
= 12.63 (not matched)
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