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). A list of ionization constants can be found here.Calculate the change in pH when 7.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

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).

Solution :-

Ka of the NH4Cl = 1*10^-14 / 1.8*10^-5 = 5.56 *10^-10

Pka = - log ka

Pka = - log 5.56*10^-10

Pka = 9.25

Now lets calculate the pH of the buffer before adding any HCl

pH= pka + log [base/acid]

pH= 9.25 + log [0.100/0.100]

pH= 9.25

now lets calculate the moles of the HCl and NH3 and NH4+

moles of HCl = 0.100 mol per L * 0.007 L = 0.0007 mol

moles of NH3 = 0.100 mol per L * 0.100 L = 0.01 mol

moles of NH4+ = 0.100 mol per L * 0.100 L = 0.01 mol

after the reaction moles of NH3 = 0.01 mol – 0.0007 mol = 0.0093 mol

moles of NH4+ = 0.01 mol + 0.0007 mol = 0.0107 mol

total volume = 0.100 L +0.007 L = 0.107 L

new molarities are as follows

[NH3] = 0.0093 mol / 0.107 L = 0.0869 M

[NH4+] = 0.0107 mol /0.107 L = 0.10 M

Now lets calculate the pH

pH= pka + log [base/ acid]

pH= 9.25 + log [0.0869/0.10]

pH= 9.19

Soo the pH change after the HCl is added = 9.19 – 9.25 = -0.06

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

Solution :-

Moles of NH3 = 0.01 mol

Moles of NH4+ = 0.01 mol

Moles of NaOH = 0.0007 mol

After the reaction

Moles of NH3 = 0.01+ 0.0007 mol = 0.0107 mol

Moles of NH4+ = 0.01 mol – 0.0007 mol = 0.0093 mol

New molarities

[NH3] = 0.0107 mol / 0.107 L = 0.10 M

[NH4+] =0.0093 mol / 0.107 L = 0.0869 M

pH= pka + log [base/acid]

pH= 9.25 + log [0.10 /0.0869]

pH= 9.31

pH change after the NaOH added = 9.31 – 9.25 = 0.06