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How many grams of NH4OH must be dissolved in water to make a 350 mL solution that has a hydroxide ion concentration of 5.2 x 10-11

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∵ NH4  on is weak base , it is partially dissociate in water.

Let say , initial concentration of NH4 OH is C

NHOH (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 NHOH , \(\alpha\) < < 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.

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