5 mL of .8 M HNO3 added to the buffer solution: 10 mL of 0.1 M CH3COOH & 20 mL of 0.2 M NaCH3COO.?
A buffer solution is made from 10 mL of 0.1 M acetic acid and 20 mL of 0.2 M sodium acetate. To the solution 5 mL of 0.8 M HNO3 is added. What is the pH of the resulting solution?( Ka for acetic acid is 1.8 × 10−5)
1. 2.51
2. 3.52
3. 0.096
4. 5.34
5. 2.79
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The HNO3 will react with the CH3COONa to produce CH3COOH and NaNO3
Therefore the concentration of CH3COOH will increase and that of CH3COONa will decrease.
Mol CH3COOH in 10.0mL of 0.1M solution = 10/1000*0.1 = 0.001 mol CH3COOH
Mol CH3COONa in 20mL of 0.2M solution = 20/1000*0.2 = 0.004 mol CH3COONa
Mol HNO3 in 5mL of 0.8M solution = 5/1000*0.8 = 0.004mol HNO3
This will react with all the CH3COONa and produce 0.004 mol CH3COOH
You now have a total of 0.001mol + 0.004 mol = 0.005 mol CH3COOH dissolved in 10+20+5 = 35mL solution
Note: you have completely destroyed the buffer and you have a solution of CH3COOH and NaNO3 - the latter is a neutral salt and does not affect the pH.
Molarity of CH3COOH solution:
Molarity = 0.005/0.035 = 0.1429 M CH3COOH solution
Determine the [H+] of the solution using the Ka equation
Ka = [H+] [CH3COO-] / [CH3COOH]
[H+] = [CH3COO-] and dissociation is small, [CH3COOH] = 0.1429M
Substitute
1.8*10^-5 = [H+]² / 0.1429
[H+]² = (1.8*10^-50*0.1429
[H+]² = 2.57*10^-6
[H+] = 1.603*10^-3
pH = -log [H+]
pH = -log ( 1.603*10^-3)
pH = 2.79
Choice 5) is correct - pH = 2.79.