Question

Question8 5 pts You add 50 g of sodium acetate (FW = 82.03) to a 1 L solution of 1 M acetic acid (pKa = 4.76). a.) what is the pH of this solution? 4.54 O 5.2 Quiz saved at 4:08pm

Answer 1

Ans. Part 1: Calculate [H₃O⁺] in 1.0 L 1M acetic acid

Given, pKa of acetic acid = 4.76

Now,

            Ka = antilog (-pKa) = antilog (-4.76) = 1.738 x 10^-5

# CH₃COOH + H₂O -----------> CH₃COO^- + H₃O⁺

Create an ICE table as shown in figure-

CH3COOHH20 > CH3coo- H30+ 1.00 Initial (mol) Chiari (mel) Equilibrium (mo1.0-X 0.00 1.00E-07 (+X) 1.00e-7X)

            Ka = [CH3COO^-] [H₃O⁺] / [CH₃COOH]       - all [Conc.] at equilibrium

Now,

            1.738 x 10^-5 = (X) (1.00 x 10^-7 + X) / (1.0 - X)

            Or, 1.738 x 10^-5 x (1.0 - X) = X² + X (1.00 x 10^-7)

            Or, 1.738 x 10^-5 – X(1.738 x 10^-5) = X² + X (1.00 x 10^-7)

            Or, X² + X(1.00 x 10^-7) + X(1.738 x 10^-5) - 1.738 x 10^-5 = 0

            Or, X² + (1.748 x 10^-5)X - 1.738 x 10^-5 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.00416            ; X2 = -0.004178

Since concertation can’t be negative, reject X2.

Hence, X = 0.00415

Therefore, [H₃O⁺] = (1.00 x 10^-7 + X) = (1.00 x 10^-7 + 0.00416) = 0.00416 M

[ Assuming 0.00416 >>1.00 x 10^-7]

# 1 mol acetic acid dissociates to yield 1 mol H₃O⁺. SO, the moles of acetic acid dissociated must be equal to the moles of H₃O⁺.

So,

[CH3COOH] dissociated = [H₃O⁺] = 0.00416 M

Now,

Remaining [CH3COOH] in solution = Initial [CH3COOH] – dissociated [CH3COOH]

                                                = 1.0 M – 0.00416 M

                                                = 0.99584 M

# Step 2: Calculate [CH3COO^-] added to initial 1L 1M acetic acid solution.

# Moles of CH₃COONa taken = Mass / Molar mass

                                    = 50.0 g/ (82.034388 g/ mol)

                                    = 0.6095 mol

Now, assuming addition of CH₃COONa cause no volume change-

            [CH₃COONa] = Moles / Volume of solution in liters

                                    = 0.6095 mol / 1.00 L

                                    = 0.6095 M

# 1 mol CH₃COONa, being a strong electrolyte, undergoes complete dissociation when dissolved in water. So, concertation of acetate ion added to the 1M acetic acid solution must be equal to the apparent concertation of sodium acetate added.

Therefore,

            [CH₃COO^-] = [CH₃COONa] = 0.6095 M

# Step 3: Calculate equilibrium [H₃O⁺] in final solution.

After addition of CH₃COONa, we have in the final solution-

            Initial [H₃O⁺] from 1M acetic acid = 0.00416 M

            Initial [CH3COOH] = 0.99584 M

            Initial [CH3COO^-] = 0.6095 M

Note that addition of CH₃COO^- to the solution shifts the equilibrium to the left following Le Chatelier’s principle.

Create an ICE table with initial concentrations mentioned above as shown in figure -

CH3COO- H30+ 0.60950 > CH3COOH+ H20 Initial (mol) Change (mol) Equilibrium (mo 0.00146 0.99584 0.6095-X 0.00416-X0.00958+X

Now,

            Keq = [CH₃COOH] / [CH₃COO^-] [H₃O⁺]

[ Note that “[CH₃COOH] / [CH₃COO^-] [H₃O⁺]” = (1 / Ka)]

            Or, 1 / Ka = [CH₃COOH] / [CH₃COO^-] [H₃O⁺]

            Or, 1 / (1.738 x 10^-5) = (0.99584 + X) / [ (0.6095 – X) (0.00416 – X)]

            Or, 57537.40 x (0.00253552 -0.61366X + X²) = 0.99584 + X

            Or, 145.89 – 35308X + 57537.40X² – 0.99584 – X

            Or, 57537.40X² – 35309X + 144.89 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.6095                          ; X2 = 0.00413

Since X can’t be greater than any of the initial concertation (because it would make one or more equilibrium concertation NEGATIVE), reject X1.

Therefore, X = 0.00413

# So, [H₃O⁺] at equilibrium = 0.00416 - X = 0.00416 - 0.00413 = 0.00003

Now,

            pH = -log [H₃O⁺] = -log (0.00003) = 4.52

# So, correct option is- B. 4.54 (closest value)