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Question
Calculate the pH of the following solution:
2 g of TlOH dissolved in water to give 2 litre of solution.
Solution
For 2g of TlOH dissolved in water to give 2 L of solution:
`["TlOH"_("aq")] = 2/2 "g/L"`
`= 2/2 xx 1/221 "M"`
`= 1/221` M
`"TlOH"_("aq") -> "Tl"_("aq")^+ + "OH"_("aq")^-`
`["TlOH"_("aq")^(-)] = ["TlOH"_("aq")] = 1/221 "M"`
`"K"_"M" = ["H"^+ ]["OH"^-]`
`10^(-14) = ["H"^+] (1/221)`
`221 xx 10^(-14) = ["H"^+]`
`=> "pH" = - log ["H"^+] = - log(221 xx 10^(-14))`
`= - log(2.21 xx 10^(-12))`
= 11.65
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