The pH of a solution containing 0-4 M HCO3- and 0-2 M CO32- is-Ka1 H2CO3 - 4 - 10-7 - Ka2 HCO3- - 4 - 10-11-Given- log 4-0-60- log 0-5-0-3

HCO_3^ → CO_3^2 + H^+ pH = pK_a_2 + log[CO_3^2 ][HCO_3^ ] = (11 log 4) + log(0.20.4 ) = 10.1 ...

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pH the Power of H

The pH of a s...

Question

The $p H$ of a solution containing $0.4 M$ $H C O_{3}^{-}$ and $0.2 M$ $C O_{3}^{2 -}$ is:
$[K_{a_{1}} (H_{2} C O_{3}) = 4 \times 10^{- 7}; K_{a_{2}} (H C O_{3}^{-}) = 4 \times 10^{- 11}]$
$(G i v e n, l o g 4 = 0.60, l o g 0.5 = - 0.3)$

12.4

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Solution

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p H

0.4 M {H C O_{3}}^{-}

0.2 M {C O_{2}}^{2 -}

K_{a_{1}} (H_{2} C O_{3}) = 4 \times 10^{- 7}; K_{a_{2}} (H C O_{3}^{-}) = 4 \times 10^{- 11}

\frac{M}{10}

K_{b}

25^{o} C i s 2.5 \times 10^{- 4},

[l o g 2 = 0.3]

p H

p H

50

0.1

H C O_{3}^{-} +

50

0.8

C O_{3}^{2 -}

H_{2} C O_{3}

K_{a_{1}} = 4 \times 10^{- 7}

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

A^{-}

A^{-}

p K_{a}

K O H

H C O O H

[K_{a} = 1.8 \times 10^{- 4}, l o g 18 = 1.26]

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