Question

What is the pH of a solution made by adding 100g of sodium acetate to 500mL of water?

please answer in detail ,

Answer 1

Solution :-

100 g Sodium acetate

500 ml water

Lets first calculate the molarity of the sodium acetate

moles = mass/ molar mass

moles of sodium acetate = 100 g / 82.0343 g per mol 1.22 mol

molarity = moles / volume in liter

Molarity of sodium acetate = 1.22 mol / 0.500 L =2.44 M

When acetate ion dissolved in water then it reacts with water to produce the OH^- ions by following equation

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

2.44 M                                      0                   0

-x                                             +x                  +x

2.44 -x                                      x                    x

Lets write the kb equation

Kb= [CH3COOH][OH^-]/[CH3COO^-]

Kb of sodium acetate is 5.55*10^-10

lets put the values in the formula

5.55*10^-10 = [x][x]/[2.44-x]

since Kb is very small therefore we can neglect the x from the denominator then we get

5.55*10^-10 = [x][x]/[2.44]

5.55*10^-10 * 2.44 = x^2

1.35*10^-9 =x^2

taking square root of both sides we get

3.67*10^-5 = x =[OH-]

now using the [OH-] lets calculate pOH

pOH= -log [OH-]

pOH = -log [3.67*10^-5]

pOH =4.44

Now lets find pH using the pOH

pH + pOH = 14

pH= 14 - pOH

pH= 14 - 4.44

pH= 9.56

Therefore pH of the solution is 9.56