If KalH2CO3=4.5×10−7M and KaHCO3−=5.0×10−11M the pH of 0.1 M Na2CO3 solution at 25 °C will be

CO32−+H2O⇌HCO3−+OH−pH=12pKw∘+pKa∘HCO3−+log⁡(c/M)=1214−log⁡5×10−11+log⁡(0.1)=12[14+10.30−1]=11.65

If KalH2CO3=4.5×10−7M and KaHCO3−=5.0×10−11M the pH of 0.1 M Na2CO3 solution at 25 °C will be

CO32−+H2O⇌HCO3−+OH−pH=12pKw∘+pKa∘HCO3−+log⁡(c/M)=1214−log⁡5×10−11+log⁡(0.1)=12[14+10.30−1]=11.65

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If $K_{al} (H_{2} {CO}_{3}) = 4.5 \times 10^{- 7} M$ and $K_{a} ({HCO}_{3}^{-}) = 5.0 \times 10^{- 11} M$ the pH of 0.1 M Na₂CO₃ solution at 25 °C will be

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10.5

8.2

9.4

11.7

answer is D.

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${CO}_{3}^{2 -} + H_{2} O ⇌ {HCO}_{3}^{-} + {OH}^{-}$

$pH = \frac{1}{2} [p K_{w}^{\circ} + p K_{a}^{\circ} ({HCO}_{3}^{-}) + \log (c / M)] = \frac{1}{2} [14 - \log (5 \times 10^{- 11}) + \log (0.1)] = \frac{1}{2} [14 + 10.30 - 1] = 11.65$

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If KalH2CO3=4.5×10−7M and KaHCO3−=5.0×10−11M the pH of 0.1 M Na2CO3 solution at 25 °C will be

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If $K_{al} (H_{2} {CO}_{3}) = 4.5 \times 10^{- 7} M$ and $K_{a} ({HCO}_{3}^{-}) = 5.0 \times 10^{- 11} M$ the pH of 0.1 M Na₂CO₃ solution at 25 °C will be