$500 m L$ of a strong acid solution of pH = 2 is mixed with $500 m L$ of another strong base solution of pOH = 3. The pH of resulting solution will be :
[Given, $log (4.5) = 0.65$ ]

3.65

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2.35

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3.35

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2.65

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Solution

The correct option is B 2.35

$p H_{s o l u t i o n 1} = 2; C_{1} = [H^{+}] = 10^{- 2} p O H_{s o l u t i o n 2} = 3; C_{2} = [O H^{-}] = 10^{- 3} C_{f} = [H^{+}]_{remains after neutralization} V_{1} = 500 m L, V_{2} = 500 m L V_{f} = V_{1} + V_{2} = 1000 m L A p p l y i n g :$
$C_{1} V_{1} - C_{2} V_{2} = C_{f} V_{f} C_{f} = \frac{C_{1} V_{1} - C_{2} V_{2}}{V_{f}} = \frac{(10^{- 2} \times 500) - (10^{- 3} \times 500)}{1000} = 4.5 \times 10^{- 3} p H = - {log}_{10} [4.5 \times 10^{- 3}] p H = 3 - 0.65 = 2.35$

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