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50 mL of 0.1 M $N a O H$ is added to $60 m L$ of 0.15 M $H_{3} {PO}_{4}$ solution $(K_{1}, K_{2} and K_{3} for H_{3} {PO}_{4} are 10^{- 3}$ , $10^{- 8} and 10^{- 13} respectively).$ The $p H$ of the mixture would be about (log 2 = 0.3) :

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6.5

5.5

3.1

4.1

answer is A.

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Detailed Solution

$H_{3} {PO}_{4} + NaOH ⟶ {NaH}_{2} {PO}_{4} + H_{2} O$

60 X 0.15 =9 5 5

4 - 5

$pH = p K_{1} + \log (\frac{5}{4})$

$= 3.1$

Hence, option A is correct.

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