First slide

PH of strong acids and strong bases

Question

The pH of a solution is 5.0. To a 10 ml of this solution, 990 ml of water is added. Then the pH of the resulting solution is

Easy

A

7
B

8
C

9
D

6.7

Solution

$V_{1} N_{1} = V_{2} N_{2}; 10 (10^{- 5}) = (10 + 990) N_{2}; N_{2} = 10^{- 7} M;$
When the concentration of the acid or base is less than equal to 10^-7 M, then [ H⁺] or [OH^-] furnished from water is also taken into accout.
$[H^{+}] = 10^{- 7} (acid) + 10^{- 7} (water); [H^{+}] = 2 \times 10^{- 7}; p^{H} = 7 - \log 2 = 6.7;$

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