Determine the degree if ionization and pH of `0.05 M` of ammonia solution. The ionization constant of ammonia can be taken from Table 7.7. Also calculate the ionization constant of the conjugate acid of ammonia.

Explanation

`K_b = ` for aniline = ` 4.27 xx 10 ^(-10) `
` C_6 H_5 NH_2 +H_2O hArr C_6H_5NH_3^(+) +OH^(-)`
` " " K_b=([C_6H_5NH_3^(+) ][OH^(-)])/([C_6H_5NH_2]) `
Since ` [C_6H_5NH_3^(+) ]=[OH^(-)]` and dissociation of aniline is not much , we have
` " " [OH^(-) ]= {K_b [C_6H_5NH_2]} ^(1//2)`
` " " = ( 4.27 xx 10 ^(-10) xx 0.001) ^(1//2)`
` therefore " " pOH = - log_(10) (6.53 xx 10 ^(-7)) = 6.185`
and ` " " pH = 14 - 6.185 = 7.815`
` {:(,C_6H_5NH_2,+, H_2O ,hArr, C_6H_5NH_3^(+) +OH^(-)),(" Initial conc. ", 0.001M ,, -,,-),(" Conc. at eq.",0.001(1-alpha),,0.001 alpha , , 0.001 alpha):} `
` " " ( because alpha ` is the degree of ionisation of aniline)
` therefore K_b= ( 0.001alpha xx 0.001alpha)/(0.001(1-alpha)) = (0.001alpha^(2))/(1-alpha) = 0.001 alpha^(2) `
` " Thus " , 0.001 alpha^(2) = 4.27 xx 10 ^(-10) `
or ` " " alpha = ((4.27 xx 10 ^(-10))/(0.001)) ^(1//2) = 6.53 xx 10 ^(-4)`
(iii) For a pair of conjugate acid and base,
` pK_b = pK_b = 14`
` pK_a= 14- pK_b = 14 -(-log_(10) 4.27 xx 10^(-10))`
` " " = 14- 9.37 = 4.63`
Hence, the ionisations constant of the conjugate acid,
` K_a = " antilog"_(10) (-pK_b) (because pK_b= - log_(10) K_b) `
` " " = " antilog"_(10) (-4.63)`
` " " = 2.34 xx 10 ^(-5)`