First slide

PH of strong acids and strong bases

Question

When 11.2 g of KOH is added to one litre of 'X' M H₂SO₄ solution, the pOH of the solution becomes 13.301. Then the value of 'X' is (assume no change in the volume of solution by the addition of KOH)

Easy

A

0.4
B

0.6
C

0.8
D

0.2

Solution

Following reaction takes place
$\large KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + {H_2}O$
$\large \left( {{n_{{H^ + }}}} \right)$
left after neutralisation = $\frac{2 \times 11.2}{56}$
Given, P^OH = 13.3
⇒ P^H = 0.7
⇒ [H⁺] = 2 $\times$ 10^-1
Thus,
$\large 2 x- \frac{{11.2}}{{56}} = 2 \times {10^{ - 1}}$
⇒ 2x = 4 $\times$ 10^-1
⇒ x = 2 $\times$ 10^-1

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