Concentration of H- ions in 0-1 M H2CO3 is K1-4-10-7-K2-4-10-11-

The correct option is A 2×10^ 4M [H^+]=√(K_1C_o)=√(4×10^ 7×0.1) =2×10^ 4M . ...

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Equivalent, Molar Conductivity and Cell Constant

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Question

Concentration of $H^{+}$ ions in 0.1 M $H_{2} C O_{3}$ is $(K_{1} = 4 \times 10^{- 7}, K_{2} = 4 \times 10^{- 11}) :$

2 \times 10^{- 4} M

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4 \times 10^{- 9} M

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2 \times 10^{- 3} M

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None of these

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Solution

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H^{+}

H_{2} S

0.1 M

H_{2} C O_{3}

0.2 M

K_{1} = 10^{- 7}, K_{2} = 10^{- 14}

H_{2} S, K_{1} = 4 \times 10^{- 7}, K_{2} = 4 \times 10^{- 11}

H_{2} C O_{3}]

0.1

N a H C O_{3}

N a_{2} H P O_{4}

N a H_{2} P O_{4}

C O_{2} + H_{2} O ⇌ H^{+} + C O_{3}^{2 -}

K_{1} = 4.2 \times 10^{- 7}

H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}

K_{2} = 4.8 \times 10^{- 11}

H_{2} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}

K_{1} = 7.5 \times 10^{- 3}

H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}

K_{2} = 6.2 \times 10^{- 8}

H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}

K_{3} = 1.0 \times 10^{- 12}

H_{2} C O_{3}

H_{2} C O_{3} ⇌ H^{+} + H C O_{3}^{-} K_{1} = 4.3 \times 10^{- 7}

H_{2} C O_{3}^{-} ⇌ H^{+} + H C O_{3}^{2 -} K_{2} = 5.6 \times 10^{- 11}

N a_{2} C O_{3}

N a H C O_{3}

K_{1} = 4.5 \times 10^{- 7}, K_{2} = 4.5 \times 10^{- 11}

K₁ and K₂ for dissociation of H₂S are 4 x 10^-3 and 1 x 10^-5. Calculate sulphide ion concentration in 0.1 MH₂S solution.

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