Question

4, Calculate the [OH1 pH and percent ionization for a 0.2M aqueous solution of NH,. K,- 1.8 x 10 5. Calculate the [H1 and pH of a buffer solution that is 0.10 M CH,COOH and 0.20 M NaCH,COO. 6. How many grams of NH.Cl are needed to 500 mL of a buffer solution that is 0.10 M NH, with a pH of 9.15?

Answer 1

4)

a)

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.2 0 0

0.2-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.2) = 1.897*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.897*10^-3 M

So, [OH-] = x = 1.897*10^-3 M

Answer: 1.90*10^-3 M

b)

use:

pOH = -log [OH-]

= -log (1.897*10^-3)

= 2.7218

use:

PH = 14 - pOH

= 14 - 2.7218

= 11.2782

Answer: 11.28

c)

% ionisation = x * 100 / initial concentration

= (1.897*10^-3)*100 / 0.2

= 0.948 %

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