A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3(aq) with 50.0 mL
ID: 1026857 • Letter: A
Question
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.
a) Calculate the NH3 concentration, the NH4Cl concentration, and the pH of the buffer solution.
b) 7.50 mL of 0.125 M HCl is added to the 100.0 mL of the buffer solution. Calculate the new NH3 concentration, NH4Cl concentration, and the pH of the buffer solution.
c) 7.50 mL of 0.125 M NaOH is added to the 100.0 mL of the original buffer solution (no HCl added). Calculate the new NH3 concentration, NH4Cl concentration, and the pH of the buffer solution.
Explanation / Answer
a)
Answer
[NH3]= 0.150M
[NH4Cl]= 0.150M
pH = 9.26
Explanation
Total volume = 100ml
dilution factor = 2
[NH3] = 0.300M/2 = 0.150M
[NH4Cl]=[NH3]= 0.300M/2 = 0.150M
Henderson-Hasselbalch equation is
pOH = pKb + log([BH+] /[B])
pKb = 4.74
BH+ = NH4+ and B = NH3
pOH = 4.74 + log(0.150M/0.150M)
pOH = 4.74
pH = 14 - pOH
pH = 14 - 4.74
pH = 9.26
b)
Answer
[NH3]= 0.1308M
[NH4+]= 0.1483M
pH = 9.21
Explanation
Added HCl react with base NH3
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 = 0.1308M
[NH4+] = (0.0159375mol/107.50ml)×1000ml = 0.1483M
applying 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
pH = 14 - 4.79 = 9.21
c)
Answer
[NH3]= 0.1483M
[NH4+]= 0.1308M
pH = 9.31
Explanation
NaOH react with NH4+
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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