Question

(a) Prove that in the aqueous solution of `NH_(4)CI` concentration of `H_(3)O^(+)` ions is `sqrt(K_(h) xx c)`.
(b) Find out the ration of `[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]]` in the aqueous solution of carbonic acid whose pH is `7.4. (K_(a) = 4.5 xx 10^(-7))`

Answer 1

(a) `NH_(4)CI +H_(2)O hArr NH_(4) OH +HCI`
The ionisation in aqueous solution is
`NH_(4)^(+) + CI^(-) + H_(2)O hArr NH_(4)OH + H^(+) + CI^(-)`
It may be wrtten as
`{:(,NH_(4)^(+),+, H_(2)O ,hArr, NH_(3) ,+, H_(3)O^(+)),("Initial molar conc :",1mol,,,,0,,0),("Equilibrium molar conc :",c(1-h),,,,c.h,,c.h):}`
(Here c = cone. of salt h = degree of hydrolysis)
Hygrolysis contant `(K_(h)) =[[NH_(3)][H_(3)O^(+)]^(+)]/((NH_(4)^(+))) =(c.hxx c.h)/(c(1-h))`
since h is very small , it can be neglected as compared to 1
`K_(h) =cg^(2) " or " h = sqrt((K_(h))/(c))`
From equilibrium reaction `[H_(3)O^(+)] =ch =c sqrt((K_(h))/(c)) = sqrt(K_(h) xx c)`
(b) it is acidic buffer
`pH =pK_(a) + log .("Conjugate base")/("Acid")`
`pH =7.4 K_(a) =4.5 xx 10^(-7) pK_(a) =7 - log 4.5 =6.35`
`pH = pK_(a) + log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] 7.4 =6.35 + log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]]`
`log .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] =7.4 -6.35 =1.05 " or " .[[HCO_(3)^(-)]]/[[H_(2)CO_(3)]] = " Antilog " 1.05 =11.22`