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_3O^+` = `(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/(35xx 10^(-3)) "mol/L" = .0428`
`pOH = -log [OH]`
= 1.36
pH = 14 - 1.36
= 12.63 (not matched)
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