Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL

Question

Calculate the change in pH when 5.00 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). A list of ionization constants can be found here.

Calculate the change in pH when 5.00 mL of 0.100 M HCI(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). A list of ionization constants can be found here. Number Calculate the change in pH when 5.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution Number H=

Explanation / Answer

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

pKb of ammonia = 4.74

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 = 5.00 x 10^-3 L x 0.100 M=0.0005

NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.0005=0.0095
moles NH4+ = 0.0100 + 0.0005=0.0105

pOH = 4.74 + log 0.0105/ 0.0095= 4.78
oH = 14 - 4.78 = 9.22

delta pH = 9.26 - 9.22 =0.04

2) no.of moles of NaOH = 0.1 X 5/1000 = 5 X 10^-4
The reaction is given by
NH4Cl + NaOH -------> NH3 + NaCl + H2O

so this time NH4Cl is consumed and NH3 is formed

new no.of moles of NH4Cl = 0.01 - 5 * 10^-4 = 0.0095
and no.of moles of NH3 = 0.01 + 5 * 10^-4 = 0.0105

new [NH4Cl] = 0.0095/0.105 = 0.0904 M
[NH3] =0.0105/0.105 = 0.1 M

pOH = 4.745 + log 0.0904/0.1

pOH = 4.745 - 0.0347 = 4.701

so new pH = 14 - 4.701= 9.299

change = 9.299 - 9.26 = 0.039

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