500.0 mL of 0.140 M NaOH is added to 625 mL of 0.200 M weak acid (Ka = 1.37 × 10-5). What is the pH of the resulting buffer?
500 mL x 0.140 = 70 millimols NaOH
625 mL x 0.200 HA = 125 mmols HA.
.......NaOH + HA ==> NaA + H2O
I......70.....125.....0.....0
C.....-70.....-70.....+70...+70
E.......0......55......70....70
Plug E line values in to HH equation and solve for pH.
pH = pKa + log(base)/(acid)
To find the pH of the resulting buffer, we need to follow these steps:
1. Calculate the moles of weak acid and NaOH:
Moles of weak acid = volume (in L) x concentration = 0.625 L x 0.200 M = 0.125 mol
Moles of NaOH = volume (in L) x concentration = 0.500 L x 0.140 M = 0.07 mol
Note: Convert mL to L by dividing by 1000.
2. Determine the amount of weak acid and conjugate base:
The weak acid reacts with the NaOH in a 1:1 ratio, meaning that 0.125 mol of weak acid reacts to form 0.125 mol of its conjugate base.
3. Calculate the total volume of the resulting solution:
The total volume can be found by adding the initial volumes of weak acid and NaOH: 0.625 L + 0.500 L = 1.125 L
4. Calculate the concentration of the weak acid and its conjugate base in the resulting solution:
Concentration of weak acid = Moles of weak acid / Total volume of solution = 0.125 mol / 1.125 L = 0.111 M
Concentration of conjugate base = Moles of conjugate base / Total volume of solution = 0.125 mol / 1.125 L = 0.111 M
5. Use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log(concentration of conjugate base / concentration of weak acid)
Given that the pKa = -log(Ka) = -log(1.37 × 10-5) = 4.86, we can substitute the values:
pH = 4.86 + log(0.111 M / 0.111 M) = 4.86 + log(1) = 4.86
Therefore, the pH of the resulting buffer is 4.86.