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JEE MainChemistryModeratePhysical Chemistry

For a solution obtained by mixing equal volumes of 0.02 M KOH
solution & 0.2 M B (weak diacidic base; = 10^-7 &
[IMAGE] = 10^-14) solution

Show answer and solution
[OH−]=0.022=0.01M∴pH=12\left\lbrack {OH}^{-} \right\rbrack = \frac{0.02}{2} = 0.01M\therefore pH = 12
10−7=0.01[BH+]0.1⇒[BH+]=10−6M10^{- 7} = \frac{0.01\left\lbrack {BH}^{+} \right\rbrack}{0.1} \Rightarrow \left\lbrack {BH}^{+} \right\rbrack = 10^{- 6}M
10−7×10−14=(0.01)2[BH22+]0.1⇒[BH22+]=10−18M10^{- 7} \times 10^{- 14} = \frac{(0.01)^{2}\left\lbrack {BH}_{2}^{2 +} \right\rbrack}{0.1} \Rightarrow \left\lbrack {BH}_{2}^{2 +} \right\rbrack = 10^{- 18}M

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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