Question

A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3(aq) with 50.0 mL...

A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3(aq) with 50.0 mL of 0.300 M NH4Cl(aq). The pKb of NH3 is 4.74.

Calculate the NH3 concentration in the buffer solution.

Calculate the NH4Cl concentration in the buffer solution.

Calculate the pH of the buffer solution.

7.50 mL of 0.125 M NaOH is added to the 100.0 mL of the original buffer solution prepared in Question 1. Calculate the new NH3 concentration for the buffer solution.
Calculate the new NH4Cl concentration for the buffer solution.
Calculate the new pH of the solution.

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Answer #1

1)

Given data,

Volume = 50.0 mL

Total volume = 50 + 50  

= 100ml

pKb = 4.74

dilution factor = 2

[NH3] = 0.300M / 2

[NH3] = 0.150M

[NH4Cl] = 0.300M / 2

[NH4Cl] = 0.150M

According to Henderson-Hassel-balch equation,

pOH = pKb + log([BH+] / [B])

BH+ = NH4+ and

B = NH3

pOH = 4.74 + log(0.150M / 0.150M)

pOH = 4.74 + log 1

pH = 14 - pOH

pH = 14 - 4.74

pH = 9.26

2)

Let us consider a reaction,

NH3 + HCl - - - - - - > NH4+ + Cl-

Initial moles of NH3 = (0.150mol / 1000ml) × 100ml

= 0.0150

Initial moles of NH4+ = (0.150mol / 1000ml) × 100ml

= 0.0150

No of moles of HCl added = (0.125mol / 1000ml) × 7.50ml

= 0.0009375

No of moles of NH3 after addition of HCl = 0.0150 - 0.0009375

= 0.0140625

No of moles of NH4+ after addition of HCl = 0.0150 + 0.0009375

= 0.0159375

Total volume = 100 + 7.50

= 107.50ml

[NH3] = (0.0140625mol / 107.50ml) × 1000ml

= 0.000130 x 1000 mL

= 0.1308M

[NH4+] = (0.0159375mol / 107.50ml) × 1000ml

= 0.1483M

According to the Henderson - Hasselbalch equation

pOH = pKb + log([BH+] /[B])

pOH = 4.74 + log(0.1483M/0.1308M)

pOH = 4.74 + 0.05

pOH = 4.79

We know pH + pOH = 14

pH = 14 - 4.79

= 9.21

3)

Let us consider a reaction,

NH4+ + OH- - - - - - - - > NH3 + H2O

No of moles of OH- added = (0.125mol / 1000ml) × 7.50ml

= 0.0009375

No of moles of NH3 after addition of NaOH = 0.0150+0.0009375

= 0.0159375M

No of moles of NH4+ after addition of NaOH = 0.0150 - 0.0009375

= 0.0140625

Total Volume = 107.50ml

[NH3] = (0.0159375mol / 107.50ml) × 1000ml

= 0.1483M

[NH4+] = (0.0140625mol / 107.50ml) × 1000ml

= 0.1308M

Applying Henderson-Hassel -balch equation

pOH = pKb + log([BH+] /[B])

pOH = 4.74 + (0.1308M/0.1483M)

pOH = 4.69

pH = 14 - 4.69

pH = 9.31

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