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The $pH$ of $\frac{M}{100}$ NaOH solution is:

(a) 2

(b) 10

(c) 6

(d) 12

1 Answer

Toolbox:
  • Ionic Product of water, $K_w = [H^+][OH^-]$
Answer: 12
 
$[OH^-] = [NaOH] = 10^{-2} M$
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-2}}= 10^{-12}$
$pH =antilog [H^+] = 12$

 

answered Nov 27, 2013 by mosymeow_1
 
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