Question

Calculate the pH of the solution that results when 15.0 mL of 0.1650 M lactic acid is A) diluted to 45.0 mL with distilled water B)mixed with 30.0 mL of 0.0825 M NaOH solution. C)mixed with 30.0 mL of 0.140 M NaOH solution. D)mixed with 30.0 mL of 0.140 M sodium lactate solution.

Answer 1

(A): Given M1 = 0.1650 M, V1 = 15.0 mL

V2 = 45.0 mL, M2 = ?

Applying law of dilution

M1V1 = M2V2

=> M2 = M1V1 / V2 = (0.1650 M x 15.0 mL) / 45.0 mL = 0.055 M

For lactic acid(C3H6O3), pKa = 3.86

=> Ka = 10^-3.86 = 1.38x10^-4

The equilibrium reaction for the dissociation of lactic acid(C3H6O3) is

-------------------C3H6O3 ----- > C₃H₅O₃^- (aq) + H⁺(aq); Ka = 1.38x10^-4

Init.Conc(M): 0.055, ------------ 0, ----------------- 0

Eqm.conc(M):(0.055 - y), ---- y, ----------------- y

Ka = 1.38x10^-4 = [C₃H₅O₃^-(aq)]x[H⁺(aq)] / [C3H6O3] = y² / (0.055 - y)

=> y = 2.69x10^-3 M

[H⁺(aq)] = y = 2.69x10^-3 M

=> pH = - log[H⁺(aq)] = - log(2.69x10^-3 M) = 2.57 (answer)

(B) The neutralization reaction is

C3H6O3 + NaOH --- > C₃H₅O₃^-(aq) + Na+(aq) + H2O

1 mol, ----- 1 mol, ----- 1 mol, 1 mol

Initial millimoles of lactic acid = MxV = 0.1650 M x 15.0 mL = 2.475 millimol

millimoles of NaOH added = MxV = 0.0825 M x 30.0 mL = 2.475 millimol

Since millimoles of NaOH = Initial millimoles of lactic acid, all of the added NaOH will be neutralized.

Hence pH of the solution is 7 (answer)

(C):

Initial millimoles of lactic acid = MxV = 0.1650 M x 15.0 mL = 2.475 millimol

millimoles of NaOH added = MxV = 0.140 M x 30.0 mL = 4.2 millimol

The neutralization reaction is

-----------------C3H6O3 + NaOH ----- > C₃H₅O₃^-(aq) + Na+(aq) + H2O

Initial mmol: 2.475, ---- 4.2, ---------- 0,

eqm.mmol:(2.475-2.475),(4.2 - 2.475), 2.475

-------------- = 0 ------------- 1.725, -------- 2.475

Hence millimoles of NaOH left = 1.725 millimol = 1.725x10^-3 mol

Total volume after reaction, Vt = 15.0 mL + 30.0 mL = 45.0 mL = 0.045 L

[NaOH] = [OH-(aq)] = 1.725x10^-3 mol /  0.045 L = 0.0383 M

pOH = - log[OH-(aq)] = - log(0.0383 M) = 1.42

=> pH = 14 - pOH = 14 - 1.42 = 12.58 (answer)

(D): millimoles of sodium lactate(C₃H₅O₃Na) = MxV = 0.140M x 30.0 mL = 4.2 millimol

Initial millimoles of lactic acid(C3H6O3) = MxV = 0.1650 M x 15.0 mL = 2.475 millimol

The solution will act as a buffer solution. Applying Hendersen equation

pH = pKa + log[C₃H₅O₃Na] / [C3H6O3] = 3.86 + log(millimol of C₃H₅O₃Na / millimol of C3H6O3)

=> pH = 3.86 + log(4.2 millimol / 2.475 millimol) = 4.09 (answer)