The document contains two chemistry problems involving buffers. The first asks which mixture would form a buffered solution, with the answer being a mixture of NH3 and NH4Cl since they can act as a weak acid and its conjugate base. The second problem involves calculating the pH of a solution containing benzoic acid and potassium benzoate after adding NaOH. The solution shows the calculation using the Henderson-Hasselbalch equation to determine the pH is 4.22.

1. 1) Which of the following mixtures would result in a buffered solution?
A) 100 mL 0.1 M NH3 plus 100 mL 0.1M NaOH
B) 100 mL 0.1 M NH3 plus 100 mL 0.1M NH4Cl
C) 100 mL 0.1 M HCl plus 100 mL 0.1M NH4Cl
D) 100 mL 0.1 M NH4NO3 plus 100 mL 0.1M NH4Cl
2) Benzoic acid, C6H5COOH is a weak acid with Ka = 6.5 x 10-5.
To 100.0 mL of buffer solution of a 0.50 M benzoic acid/0.50 M potassium benzoate 10. mL of
0.20 M NaOH is added. What is the pH of the resulting solution?
Solution
1)
Answer : B) 100 mL 0.1 M NH3 plus 100 mL 0.1M NH4Cl
Explanation :
buffer is formed from , weak acid and its conjugate base or weak base and its conjugate acid.
here A) NH3 and NaOH both are bases. so cannot form a buffer.
C) HCl and NH4Cl , these cannot form a buffer.
D) NH4NO3 and NH4Cl both are salts so these cannot form a buffer.
2)
Ka = 6.5 x 10^-5
pKa = 4.187

2. mmoles of acid = 100 x 0.50 = 50
mmoles of salt = 100 x 0.50 = 50
mmoles of NaOH = 10 x 0.20 = 2
pH = pKa + log [salt + C / acid - C]
= 4.187 + log [50 + 2 / 50 - 2]
pH = 4.22