When 200 mL solution of HCl of p^H=2 is mixed with 300 mL solution of NaOH p^H=12, the pH of resulting solution is (log2 = 0.3)

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2.7

11.3

answer is B.

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Detailed Solution

$H C l + N a O H \to N a C l + H_{2} O$

pH = $- l o g [H^{+}]$ , $m o l e s = c o n c e n t r a t i o n \times v o l u m e$

Explanation: For HCl, pH = 2

pH = $- l o g [H^{+}]$

2 = $- l o g [H^{+}]$

$[H^{+}]$ = $10^{- 2}$ = 0.01M

$m o l e s = c o n c e n t r a t i o n \times v o l u m e$ = 0.01 x 200 =2

Similarly, for NaOH pH =12 then pH + pOH = 14

12 + pOH = 14

pOH = 14-12 = 2

pOH = $- l o g [O H^{-}]$

2 = $- l o g [O H^{-}]$

$[O H^{-}] = 10^{- 2}$ = 2

moles = 0.01 x 300 = 3

Total volume = 200 + 300 = 500ml

Out of 3 moles of NaOH, 2 moles of NaOH will be neutralized with 2 moles of HCl (3-2=1).

the $[O H^{-}]$ concentration remaining = $\frac{1}{500}$ = $2 \times 10^{- 3}$ M

pOH = $- l o g [O H^{-}]$ = $- l o g [2 \times 10^{- 3}]$ = 2.70

Then pH + pOH = 14

pH = 14 - 2.70 = 11.3.

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