Question : Which will make basic buffer ?
(1) 50 mL of 0.1M NaOH + 25 mL of 0.1 M CH3COOH
(2) 100 mL of 0.1M CH3COOH + 100 mL of 0.1M NaOH
(3) 100 mL of 0.1M HCl + 200 mL of 0.1M NH4OH
(4) 100 mL of 0.1M Hcl + 100 mL of 0.1M NaOH
answer : from all above options the correct answer is “(3) 100 mL of 0.1M HCl + 200 mL of 0.1M NH4OH”
Explanation of correct answer :
Out of the given options , NaOH is a strong base , so it’s conjugate acid is a weak. Similarly , HCl is a strong acid so it’s conjugate base is weaker. CH3COOH is a weak acid , so its conjugate base is stronger. NH4OH is a weak base , so its conjugate acid is stronger.
NaOH + CH3COOH → CH3COONa + H2O
NaOH + HCl → NaCl + H2O
NH4OH + HCl → NH4Cl + H2O
So , 1 mol of NaOH reacts with 1 mol of CH3COOH
For (1) : 50/1000 x 0.1 mol of NaOH reacts with
25/1000 x 0.1 M of CH4COOH
As number of moles of CH3COOH is less , it is wholly consumed in the reaction. So , unreacted NaOH = 0.005 – 0.0025 = 0.0025 mol
For salt hydrolysis of weak acid (CH3COOH) and strong base (NaOH) , we have
pH = ½ [pKw + pKa + log [CH3COO–]]
Vtotal = 50 + 25 = 75 mL.
[CH3COO–] = 0.0025 x 75 / 1000 = 0.0001875 M
So
pH = 7 + pKa/2 + log(0.0001875) = 7 + pKa/2 – 3.72 = 3.27 + pKa/2
So , it forms an acidic buffer.
For option (2) : 100/1000 x 0.1 mol of NaOH reacts with
100/1000 x 0.1 M of CH3COOH
As number of moles of NaOH and CH3COOH are equal , both are wholly consumed in this reaction.
For salt hydrolysis of weak acid (CH3COOH) and strong base (NaOH) , we have
pH = ½ [pKw + pKa + log [CH3COO–]]
Vtotal = 100 + 100 = 200 mL
[CH3COO–] = 0.01 x 200 / 1000 = 0.002M
So , pH = 7 + pKa/2 + log (0.002) = 7 + pKa/2 – 2.69 = 4.3 + pKa/2
So , it forms and acidic buffer.
For option (3) 200/1000 x 0.1 mol of NH4OH reacts with
100/1000 x 0.1 M of HCl
As number of moles of HCl is less , it is wholly consumed in the reaction. So unreacted NH4OH = 0.02 – 0.01 = 0.01 mol
For salt hydrolysis of strong acid (HCl) and weak base (NH4OH) , we have
pH = ½ [pKw + pKb – log [NH4Cl]]
Vtotal = 200 + 100 = 300 ml
[NH4Cl] = 0.01 x 300 / 1000 = 0.003 M
So , pH = 7 + pKb/2 – log (0.003) = 7 + pKb/2 + 2.69 = 9.69 + pKb/2
So , it forms a basic buffer.
For option (4) : 100/1000 x 0.1 mol of NaOH reacts with
100/1000 x 0.01 M of HCl
As number of moles of HCl and NaOH are equal , they completely reacts with each other and no buffer is formed.
extra information : buffer is a type of solution that contains acidity or we can say alkalinity and this type of solution will oppose the change of pH when some acid or alkali is mixed with it in small amount.
option (1) : this is an example of basics solution not an example of basic buffer solution. so this is incorrect answer.
option (2) : in this reaction Hydrolysis of salt happened but not basic buffer solution so this is also incorrect answer.