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Question
If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Solution
`["KOH"_("aq")] = 0.561/(1/5) "g"/"L"`
= 2.805 g/L
`= 2.805 xx 1/56.11 "M"`
= 0.05 M
`"KOH"_("aq") -> "K"_("aq")^+ + "OH"_(("aq"))^-`
`["OH"^-] = 0.05 "M" = ["K"^+]`
`["H"^+] ["H"^-] = ["K"^+]`
`["H"^+] "K"_"w"/(["OH"^-])`
`= 10^(-14)/0.05 = 2xx 10^(-13) "M"`
`therefore "pH" = 12.70`
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