Question

You have 7 liters of solution at a PH of 6.7. Every day 35 grams of Acetic Acid (CH3COOH) is added to the solution. For every one mol of Acetic Acid added, 2 mols of H2O are also added. What is the PH of the solution after 10 days, 20 days, and 30 days?

Answer 1

Vinitial = 7 L

pH = 6.7

day 0 = 0

day 10 = 35*10 = 350 g of acetic acid

mol = mass/MW = 350/60 = 5.8333333

so rati ois 1:2 so

5.8333333 mol of Acid = 5.8333333*2 = 11.6666666 mol of water

mass = mol*MW = 11.6666666*18 = 209.9 mL = 0.2091 L

Vtotal = 7+0.2091 = 7.2091 L

[HA] = mol/V = 5.8333333 / (7.2091) = 0.809162

Ka = [H+][A-]/[HA]

1.8*10^-5 = x*x/(0.809162-x)

x = 0.0038

pH = -log(0.0038) = 2.4202

b)

d = 20 days (2x 1st case)

mass = 35*20 = 700 g

mol = 5.8333333*2 = 11.6666666 mol of acid

mol of H2O = 11.6666666*2 = 23.3333

mass = 23.33*18 = 419.94 g = 419.94mL = 0.41994 L

Vtotal = 7+0.41994 = 7.41994 L

[HA] = 11.6666666 /7.41994 = 1.5723

so

Ka = [H+][A-]/[HA]

1.8*10^-5 = x*x/(1.5723-x)

[H+] = 0.0053

pH = -log(0.0053) = 2.2757

finally

c)

d = 30 days (3x 1st case)

mass = 35*30 = 1050 g

mol = 5.8333333*3 = 17.4999 mol of acid

mol of H2O = 17.4999*2 = 34.9998

mass = 34.9998*18 = 629.9964 g = 629.9964 = 0.6299964 L

Vtotal = 7+0.6299964 = 7.6299964 L

[HA] = 17.4999 /7.6299964= 2.29356

so

Ka = [H+][A-]/[HA]

1.8*10^-5 = x*x/(2.29356-x)

[H+} = 0.0064

pH = -log(0.0064) = 2.193