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7.49     Calculate the pH of the following solutions:

  (c)     0.3 g of NaOH dissolved in water to give 200 mL of solution. 

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NaOH dissociates into NaOH\rightarrow Na^+ +OH^-
So, the concentration of [NaOH]  = \frac{0.3\times 1000}{200\times 40} M = 0.0375 M

\therefore [OH^-] = [NaOH] = 0.0375 M
 

We know that ,
[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.0375} = 2.66\times 10^{-13}

Now, P^H = - \log[H^+]
                  \\=- \log(2.66\times 10^{-13})\\ =12.57

Posted by

manish

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