1. (20 points) An analytical balance was used to weigh 200.01 mg o solid sodium

Question

1. (20 points) An analytical balance was used to weigh 200.01 mg o solid sodium acetate(NaCJH3o-NaAc MW = 82.0343, pKa (HC-H3O2 ~ HAc)- 4.763 which was transferred to a 100.00 mL volumetric flask that was then filled to the line with carbonate-free water and mixed well. Determine the pH of each of the solutions below. f d ry A.) The original sodium acetate solution B.) A 25.00 mL aliquot of the NaAc solution plus 25.00 mL of 0.100 M aqueous acetic acid (HAc) solution C.) A 25.00 mL aliquot of the NaAc solution plus 15.00 mL of 0.100 M aqueous hydrogen chloride solution

Explanation / Answer

sodium aceate is base formed from reaction of weak acetic acid (Pka= 4.76) , Ka= 10(-4.76)= 1.74*10-5 and strong base NaOH. moles of sodium acetate= mass in gm/ molar mass, molar mass of CH3COONa= 82.043 g/mole

1000 mg= 1gm and 1000ml= 1L

moles of sodium acetate= 200mg*10-3 g/mg/ 82.043 =0.00243

volume of solution =100ml. volume of solution in liters= 100/1000=0.1L

concentration of sodium acetate=0.00243/0.1=0.0243M

sodium acetate (CH3COONa) ionizes as CH3COO- + H2O -------->CH3COOH+OH-

Kb= 10-14/Ka= 10-14/ (1.74*10-5)= 5.88*10-10 = [CH3COOH][OH-]/[CH3COO-]

let x= drop in concentration of sodium acetate to reach equilibrium

at equilibrium, [OH-]= [CH3COOH]= x and [CH3COO-]=0.0243-x

Kb= x2/(0.0243-x)= 5.88*10-10, when solved for x, x=3.78*10-6, pOH= -log [OH-]= 5.42, pH= 14-pOH= 14-5.42=8.58

2. moles of sodium acetate= molarity* volume in L= 0.0243*25/1000 = 0.00061

moles of acetic acid = 0.1*25/1000=0.0025, volume of solution after mixing = 25+25= 50ml= 50/1000L

concentrations: sodium acetate = moles/volume in liters= 0.00061/(50/1000) and acetic acid =0.0025/(50/1000)

From Henderson-Hasselbalch equation, pH= pKa+ log {[sodium acetate]/[Acetic acid]} (1)

substituting the value in Eq.1 gives pH= 4.76+log (0.00061/0.0025)= 4.14

when 15m of 0.1M HCl is added, moles of HCl added = 0.1*15/1000=0.0015

HCl reacts with sodium acetate as follows CH3COONa+ HCl ------->CH3COOH+ NaCl. acetic acid is formed

the molar ratio of CH3COONa: HCl ( theoretical)= 1:1 actual molar ratio = 0.00061:0.0015

so excess is HCl and all the sodium acetate reacts. moles of HCl remaing = 0.0015-0.00061= 0.00089

volume of solution after mixing = 25+15= 40ml

volume of solution in liters= 40/1000, concentration of HCl remaining = 0.00089/(40/1000)= 0.02225

since HCl is strong acid and ionizes completely as HCl+ H2O -------->H3O++Cl-

[H3O+]=0.02225, pH=-log (H3O+)= 1.65

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