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`        PH of solution containing 0.2M H2CO3 and 0.1M CO3 2- is (for  H2CO3  ka 1 =4*10-7and ka 2=4*10-11.) `
5 months ago

Vikas TU
10059 Points
```							Dear student Dissociation constant of n acid is the rati of its molar concentrations [H+][A-]/[HA]=KaOn the case of your question,[H+][CO3^2-]/[HCO3-]=4×10^(-7) , so[H+]=([HCO3-]/[CO3^2-])×4×10^(-7)=2.0×4×10^(-7)=8×10^(-7) .pH=-log([H+])=-log(8.0×10^(-7)) => aproximately equal to 6.40
```
5 months ago
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• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions