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Acids and Bases in Water

At 25oC, wh...

Question

At $25^{o} C$ , what is the $p O H$ (nearest integer) when $0.1$ millimole of a strong acid is added to $25 m L$ of $0.025 M$ $N a O H$ ?

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Solution

First calculate the pOH of 0.025 M NaOH solution.
$p O H = - l o g [O H^{-}] = - l o g 0.025 = 1.602$ .
Now calculate its pH.
$p H = 14 - p O H = 14 - 1.602 = 12.398$
Now calculate the number of millimoles of NaOH solution.
$0.025 M \times 25 m L = 0.625 m i l l i m o l e s$
Out of this, 0.1 millimoles will be neutralized by the acid.
Hence, the number of millimoles of NaOH remaining is $0.625 - 0.1 = 0.525$
The hydroxide ion concentration is $\frac{0.525 m i l l i m o l e s}{25 m l} = 0.021 M$
Now calculate the pOH of the solution.
$p O H = - l o g [O H^{-}] = - l o g 0.021 = 1.677$ .
Now calculate the pH of the solution.
$p H = 14 - p O H = 14 - 1.677 = 12.322$
$p O H = 14 - p H = 14 - 12.322 \approx 2$

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