1.

Calculate the hydrolysis constant (ii) degree of hydrolysis (iii) pH of 0.1M CH_3COONa solution (pK_a for CH_3COOH is 4.74)

Answer»

Solution :`CH_3COONa` is a salt of weak acid `(CH_3COOH)` and a strong base (NaOH).
Hence, the solution is ALKALINE due to hydrolysis.
`CH_3COO^-(aq)+H_2O (aq) leftrightarrow CH_3COOH(aq)+OH^-(aq)`
Given that `pK_a=4.74`
`K_a`= antilog of `(-pK_a)`
=antilog of (-4.74)
=antilog of (-5+0.26)
`=10^-5 times 1.8`
[antilog of `0.26=1.82 APPROX 1.8`]
`h=sqrt((K_w)/(K_a times c))=sqrt((1times10^-14)/(1.8 times 10^-5 times0.1))=7.5 times 10^-5`
(II) `K_h=K_w/K_a=(1 times 10^-14)/(1.8 times10^-5)=5.56 times 10^-10`
(iii) `pH=7+(pK_a)/2+(LOG c)/2=7+4.74/2+log0.1/2=7+2.37-0.5=8.87`


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