The OH - concentration in a mixture of 5 . 0 mL of 0 . 0504 M NH 4 Cl and 2 mL of 0 . 0210 M NH 3 solution is x × 10 - 6 M . The value of x is (Nearest integer) [Given K w = 1 × 10 - 14 and K b = 1 . 8 × 10 - 5

NH 4 Cl = NH 4 + = 0 . 0504 M V = 5 . 0 mL NH 3 = 0 . 0210 M V = 2 . 0 mL K b = NH 4 + OH - NH 3 OH - = K b × NH 3 NH 4 + = 1 . 8 × 10 - 5 × 0 . 0210 × 2 0 . 0504 × 5 = 3 × 10 - 6 M

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesMedium2 minPYQ_2021

ChemistryMediumnumerical

The ${OH}^{-}$ concentration in a mixture of $5.0 mL$ of $0.0504 M {NH}_{4} Cl$ and $2 mL$ of $0.0210 M {NH}_{3}$ solution is $x \times 10^{- 6} M$ . The value of $x$ is (Nearest integer) [Given $K_{w} = 1 \times 10^{- 14}$ and $K_{b} = 1.8 \times 10^{- 5}]$

Answer:

3.00

Solution:

$[{NH}_{4} Cl] = [{NH}_{4}^{+}] = 0.0504 M V = 5.0 mL$ $[{NH}_{3}] = 0.0210 M V = 2.0 mL$

$K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{3}]}$

$[{OH}^{-}] = \frac{K_{b} \times [{NH}_{3}]}{[{NH}_{4}^{+}]} = \frac{1.8 \times 10^{- 5} \times 0.0210 \times 2}{0.0504 \times 5} = 3 \times 10^{- 6} M$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Strong and Weak Acids and Bases

⏱ 2mℹ️ Source: PYQ_2021

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