∵ NH4 on is weak base , it is partially dissociate in water.
Let say , initial concentration of NH4 OH is C
NH4 OH (aq)
∴ Kb = \(\cfrac{[NH_4^+][\bar OH]}{[NH_4OH]}\)
Kb = \(\cfrac{C\alpha \times c\alpha}{c(1-\alpha)}\)
Kb = \(\cfrac{C\alpha^2}{(1-\alpha)}\)
for NH4 OH , \(\alpha\) < < i
∴ 1 - \(\alpha\) ≈ 1
kb = \(C\alpha ^2\) ----------------- (1)
We have given,
concentration \(\bar{O}H\) = C \(\alpha\) = 5.2 x 10-11
∴ \(\alpha\) = \(\cfrac{5.2\times10^{-11}}{C}\) ------------ (2)
Putting the value of \(\alpha\) in equation (1)
Kb = c x \(\cfrac{(5.2\times10^{-11})}{C}^2\)
1.8 x 10-5 = \(\cfrac{(5.2\times10^{-11})}{C}^2\)
C = \(\cfrac{(5.2\times10^{-11})}{1.8\times10^{-5}}^2\)
C = \(\cfrac{27.04\times10^{-22}}{1.0\times10^{-5}}\)
C = 15.02 x 10-17 M
Let say . m gram of NH4 OH is dissolve in 350ml water.
Then
15.02 x 10-17 = \(\cfrac{(\frac{M}{35})\times10000}{350}\)
M = \(\cfrac{15.02\times10^{-17}}{100}\)
M = 15.02 x 10-15 g.