Please explain why the following dilutions pH behaves the way that they do: a) (
ID: 833448 • Letter: P
Question
Please explain why the following dilutions pH behaves the way that they do:
a) (0.1 M CH3COOOH and 0.1 M CH3COONa)
Initial Ph: 4.75
1 to 10 dilution: 4.75
Addition of 1.0 mL of 0.1 M HCl to 100 mL of solution: 4.75
Addition of 1.0 mL of 0.1 M NaOH to 100 mL of solution : 4.77
b) (0.1 M CH3COOH and 0.01 M CH3COONa)
Initial Ph: 3.76
1 to 10 dilution: 3.76
Addition of 1.0 mL of 0.1 M HCl to 100 mL of solution: 3.70
Addition of 1.0 mL of 0.1 M NaOH to 100 mL of solution: 3.80
c) (0.01 M CG3COOH and 0.1 M CH3COONa)
Initial Ph: 5.76
1 to 10 dilution: 5.76
Addition of 1.0 mL of 0.1 M HCl to 100 mL of solution: 5.71
Addition of 1.0 mL of 0.1 M NaOH to 100 mL of solution: 5.80
Explanation / Answer
these are buffer
for buffer we used henderson equation
this one is the strong base and weak acid case
pH = pKa +log[salt/acid ]
if we add strong acid then it resist ro increase pH of solution .and try to reduce to the pH and decrease the molarity of salt
if we add strong base then it will increase the molarity of salt . thay why there is some increament in pH after adding some strong acid