Question

12) Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

delta pH=?

Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

delta pH=?

Answer 1

A)

so you are adding a strong acid to your buffer which will reduce the base concentration and raise your acid concentraion by 3.00E-4 mole

so now your concentrations are

(0.010-3.00E-4) mole=NH3
(0.010+3.00E-4) mole=NH4Cl

9.7E-3 mole=NH3
1.03E-2 mole=NH4Cl

Now you need the Ka value which is found by dividing the Kw value(1.00E-14) y the Kb value for NH3 (1.8E-5)

Ka=1.00E-14/1.8E-5
Ka=5.5556E-10 now take the nagative log to find the pKa value
pKa=9.2553

now use all these numbers to find the pH using the henderson-Hasselbalch equation
pH=pKa+log([base]/[acid])
pH=9.2553+log(9.7E-3 mole/1.03E-2 mole)
pH=9.229

the original pH is just the pKa value since you had equal amounts of NH3 and NH4Cl so the difference is

delta pH=9.2553-9.229
delta pH=(0.0263) pH units

B)  

no.of moles of NaOH = 0.1 X 3/1000 = 3 X 10^-4

this time this reaction will take place :-

NH4Cl + NaOH -------> NH3 + NaCl + H2O

so this time NH4Cl is consumed and NH3 is formed

new no.of moles of NH4Cl = 0.01 - 3 X 10^-4 = 0.0097
and no.of moles of NH3 = 0.01 + 3 X 10^-4 = 0.0103


pOH = 4.745 + log 0.0097/0.0103

pOH = 4.7189

so new pH = 14 - 4.7189 = 9.281

change = 9.281 - 9.255 = 0.026