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# Calculate the pH of the following solutions: (c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

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

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