FeOH2 is a diacidic base with Kb1 - 10-4 and Kb2 - 2-5-10-6- What is the concentration of FeOH2 on 0-1 M FeNO32 solution-

The correct option is B 10^ 10 The equilibrium reaction is as shown below. Fe(OH)_2 #160; ⇌ Fe^2+ + 2OH^ The equilibrium constant is K=K_a1× k ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Acid

FeOH2 is a di...

Question

$F e (O H)_{2}$ is a diacidic base with $K_{b_{1}} = 10^{- 4}$ and $K_{b_{2}} = 2.5 \times 10^{- 6}$ . What is the concentration of $F e (O H)_{2}$ on $0.1 M F e (N O_{3})_{2}$ solution?

4 \times 10^{- 9}

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2.5 \times 10^{- 6}

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10^{- 10}

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10^{- 14}

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Solution

The correct option is B $10^{- 10}$
The equilibrium reaction is as shown below.
$F e (O H)_{2} ⇌ F e^{2 +} + 2 O H^{-}$
The equilibrium constant is $K = K_{a 1} \times k_{a 2} = 1 \times 10^{- 4} \times 2.5 \times 10^{- 4} = 2.5 \times 10^{- 10}$
The expression for the equilibrium constant $= K = \frac{[F e^{2 +}] [O H^{-}]^{2}}{[F e (O H)_{2}]}$
Substitute values in the above expression, we get
$2.5 \times 10^{- 10} = \frac{(0.1 - x) (1.0 \times 10^{- 7} - 2 x)^{2}}{x}$
On solving this equation, we get $x = 1 \times 10^{- 10}$

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Fe(OH)2 is a diacidic base with Kb1=10−4 and Kb2=2.5×10−6. What is the concentration of Fe(OH)2 on 0.1 M Fe(NO3)2 solution?

$F e (O H)_{2}$ is a diacidic base with $K_{b_{1}} = 10^{- 4}$ and $K_{b_{2}} = 2.5 \times 10^{- 6}$ . What is the concentration of $F e (O H)_{2}$ on $0.1 M F e (N O_{3})_{2}$ solution?