Question

What is the pH of a buffer solution made by dissolving 10 g of sodium acetate in 200 ml of 1M acetic acid? The Ka for acetic acid id 1.7*10^-5.

B. Calculate the pH of the solution if 10 ml of 0.100M HCl is added to the solution.

C. What will the pH be if 5.00 ml of 0.150M NaOH was added to solution in B?

D. Write the equation when an acid and a base are added to the buffer system.

Answer 1

What is the pH of a buffer solution made by dissolving 10 g of sodium acetate in 200 ml of 1M acetic acid? The Ka for acetic acid id 1.7*10^-5.

Solution :-

Lets first calculate the moles of the sodium acetate

Moles = mass/ molar mass

Moles of sodium acetate= 10 g /82.0343 g per mol = 0.122 mol

Molarity of the sodium acetate = 0.122 mol / 0.200 L = 0.610 M

Pka of acetic acid = -log ka

Pka= -log 1.7*10^-5

Pka= 4.77

Now lets calculate the pH using the Henderson equation

pH= pka+ log [base/acid]

pH= 4.77 + log [0.61/1]

pH= 4.56

B. Calculate the pH of the solution if 10 ml of 0.100M HCl is added to the solution.

Solution :-

Moles of acetic acid = 1.0 mol per L * 0.200 L = 0.200 mol

Moles of sodium acetate = 0.122 mol

Moles of HCl = 0.100 mol per L * 0.010 L = 0.001 mol

After the reaction moles of acetic acid = 0.200 mol + 0.001 mol = 0.201 mol

Moles of sodium acetate = 0.122 mol – 0.001 mol = 0.121 mol

Total volume = 200 ml + 10 ml = 210 ml = 0.210 L

New molarity of the acetic acid = 0.201 mol / 0.210 L =0.9571 M

New molarity of the acetate = 0.121 mol / 0.210 L = 0.5762 M

pH= pka + log [base/acid]

pH= 4.77 + log [0.5762 /0.9571]

pH= 4.55

C. What will the pH be if 5.00 ml of 0.150M NaOH was added to solution in B?

Solution :-

Moles of NaOH = 0.150 mol per L * 0.005 L = 0.00075 mol

New moles of acetic acid = 0.200 mol – 0.00075 mol = 0.19925 mol

New moles of sodium acetate = 0.122 mol + 0.00075 mol = 0.12275 mol

New molarity at the total volume is

Molarity of the acetic acid = 0.19925 mol / 0.205 L = 0.972 M

Molarity of the acetate = 0.12275 mol / 0.205 L = 0.59879 M

pH= pka+ log [base/acid]

pH= 4.77 + log [0.59879/0.972]

pH= 4.56

D. Write the equation when an acid and a base are added to the buffer system

Solution :-

Equation when acid is added (added acid reacts with the acetate ion to form the acetic acid)

CH3COO^- + H+ ------- > CH3COOH

Equation when base is added (Added base reacts with acetic acid to form the acetate ion)

CH3COOH + OH-   --------- > CH3COO^ -   + H2O