equilibrium acid and base buffer?

Question

plz someone help me on this question !
i would be appreciated ! ^.^

NH4+ is a weak acid whose Ka=5.6x10^-10. Determine the following :

a.The Kb for a solution of NH3.

b.The pH of a 200 ml of a 0.5 M solution of NH3

c. 50 ml of a 0.5 M solution of HCl is then added to the solution of NH3. Determine the pH of the resulting solution.

d. Determine the pH of the solution when a total of 100 ml of the HCl is added.

e. Determine the pH of the solution when a total of 200 ml of
the HCl is added.

f. Determine the pH of the solution when a total of 800 ml of the HCl is added.

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i only know how to get part a, the rest of parts r confusing ...

Answer 1

a) Kb = Kw / Ka = 1.78 x 10^-5

b ) NH3 + H2O <> NH4+ + OH-
initial concentration
......0.5 . . . . . . . . . . .. 0 . . . . .0

at equilibrium
. . 0.5-x . . . . . . . . . . . .x . . . . .x

K = 1.78 x 10^-5 = x^2 / 0.5-x

x = 0.00299 M

pOH= 2.5

pH = 11.5

c) Moles NH3 = 0.5 ( 200 ) /1000 = 0.10
moles HCl = 0.5 (50)/1000= 0.025

Total volume = 250 mL = 0.250 L

0.10 mole NH3 + 0.025 mole H+ >> 0.025 mole NH4+ + 0.075 mole NH3 in excess

Concentration NH4+ = 0.025 / 0.250= 0.1 M
Concentration NH3 = 0.075 / 0.250 = 0.30 M

1.78 x 10^-5 = (0.1 + x)(x) / 0.30-x

x = 0.0000534 M

pOH= 4.27 >> pH = 9.73


d) moles NH3 = 0.10 M

moles HCl = 0.5 (100)/1000= 0.05

we get 0.05 mole NH3 = mole NH4+

Total volume = 300 mL

0.05 / 0.3 = 0.166 M

1.78 x 10^-5 = 0.166+x(x)/0.166-x

x= 1.78 x 10^-5

pOH= 4.75
pH= 9.25


e) moles NH3 = 0.10 M
moles HCl = 0.5(200)/1000=0.10

we get 0.10 mole NH4+
[NH4+] = 0.10 / 0.400 = 0.25 M

NH4+ <> NH3 + H+

Ka= 5.6 x 10^-10 = x^2 / 0.25-x

x = 1.18 x 10^-5 M

pH=4.93


f) moles NH3 = 0.10

moles HCl = 0.5(800)/1000= 0.4

we get 0.3 mole H+ left in excess plus 0.1 mol NH4+

Volume = 1 L

[H+] = 0.1 M

pH = 1

Answer 2

at the same time as the acid and its conjugate base are both modern-day, you've a buffer. If the acid and the conjugate base have a similar concentration, it really is a particular case. Then the pH equals the pKa of the acid. Take the log (base 10) of one million.8E-05 and take the unfavorable of that.

Answer 3

a) pKb = 14-pKa

b) pH = -log(0.5)