1.

Calcualte [OH^(-)] & [H_(2)C_(2)O_(4)] in a 0.005M Na_(2)C_(2)O_(4) solution.Given :K_(a_(1))&K_(a_(2)) for oxalic acid are 5.6xx10^(-2)& 5.4xx10^(-5)Take sqrt((1)/(108))=0.096

Answer»

`[OH^(-)]=9.6xx10^(-7)M`
`[H_(2)C_(2)O_(4)]=(5)/(28)xx10^(-12)M`
Both (A) and (B)
None of these

Solution :`{:(,C_(2)O_(4)^(2+)+H_(2)OhArr,HC_(2)O_(4)^(-),+,OH^(-),,K_(h_(1))=(K_(w))/(K_(a_(1)))=(10^(-12))/(5.6)),(t=0,0.005(1-H),0.005h,,0.005h,),(,,~~0.005,,,):}`
`((0.005h)(0.005h))/((0.005))=(10^(-9))/(5.4):.h=1.92xx10^(-4)`
`[OH^(-)]=0.005h=9.6xx10^(-7)M.`
`{:(,HC_(2)O_(4)^(-)+H_(2)OhArr, H_(2)C_(2)O_(4),+,OH^(-),,K_(h_(2))=(K_(w))/(K_(a_(1)))=(10^(-12))/(5.6)),(t=eq,~~9.6xx10^(-7),,,~~9.6xx10^(-7),):}`
`[H_(2)C_(2)O_(4)]=K_(h_(2))=(5)/(28)xx10^(-12)M`


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