1.

Calculate the i) Hydrolysis constant,

Answer»

Solution :Sodium carbonate is a salt of weak ACID, `H_2 CO_3` and a strong base, `NaOH`, and hence the solution is ALKALINE due to hydrolysis.
`Na_2 CO_(3(aq)) to 2Na_((aq))^(+)+CO_(3(aq))^(2-)`
`CO_(3(aq))^(2-)+H_2O(I) HARR HCO_(3)^(-)+OH^(-)`
(i) `h=sqrt((K_w)/(K_axxC))`
`=sqrt((1XX10^(-14))/(1.8xx10^(-5)xx0.1))`
`h=7.5 xx10^(-5)`
Given that `pK_a=4.74`
`pK_a=-log K_a`
i.e. , `K_a=` ANTILOG of `(-pK_a)`
= antilog of `(-4.74)`
= antilog of `(-5+0.26)`
`=10^(-5)xx1.8`
[ antilog of `0.26=1.82 cong 1.8`]


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