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7.49     Calculate the pH of the following solutions:

 (a)     2 g of TlOH dissolved in water to give 2 litre of solution.

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Here, 2 g of TlOH dissolves in water to give 2 litres of solution
So, the concentration of   TlOH = 
[TlOH(aq)] = \frac{2}{2}g/L = \frac{1}{221}M     (the molar mass of  TlOH is 221)

TlOH can be dissociated as TlOH(aq)\rightarrow Tl^{+}+OH^-

OH^-(aq)= TlOH(aq)=\frac{1}{221}M

Therefore, K_w = [H^+][OH^-]            (since Kw = 10^{-14})

So, the concentration of [H^+] = 221\times 10^{-14}

Thus P^H = -\log[H^+] = -\log(221\times 10^{-14})
                                               = 11.65(approx)

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