Question

Calculate the change in pH when 7.00 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).

Calculate the change in pH when 7.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

For the first case, when 7 mL of 0.1 M HCl is added to buffer,

We know,

pOH = pKb + log [NH4+]/ [NH3]

pKb of ammonia = 4.74 (standard value)

initial pOH = 4.74 + log 0.100/0.100 = 4.74

pH = 14 - 4.74=9.26

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100

moles H+ added = 0.007 L x 0.100 M=0.0007

Now, NH3 + H+ = NH4+

moles NH3 = 0.0100 - 0.0007 = 0.0093

moles NH4+ = 0.0100 + 0.0007 = 0.0107

pOH = 4.74 + log 0.0107/ 0.00930= 4.80

pH = 14 - 4.80 = 9.20

Therefore, change in pH after adding 7 mL of 0.1 M HCl = 9.26 - 9.20 =0.06

In the second case 7 mL of 0.1 M NaOH is added to the buffer. Using the previous given equation,

pOH = pKb + log [NH4+]/ [NH3]

pKb of ammonia = 4.74 (standard value)

initial pOH = 4.74 + log 0.100/0.100 = 4.74

pH = 14 - 4.74=9.26

moles NH3 = 0.100 L x 0.100 M = 0.0100

moles OH- added = 0.007 L x 0.100 M=0.0007

Now, NH4+ + OH- = NH3 + H2O

moles NH3 = 0.0100 + 0.0007 = 0.0107

moles NH4+ = 0.0100 - 0.0007 = 0.0093

pOH = 4.74 + log 0.0093/ 0.0107= 4.68

pH = 14 - 4.68 = 9.32

Therefore, change in pH after adding 7 mL of 0.1 M NaOH = 9.26 - 9.32 = -0.06