Question

Prepare a buffer solution with: 0.5 g NaC2H3O3 + 2 mL of 3 M acetic acid + 12 mL water Use 5 mL buffer + 0.2 mL 6 M NaOH What is the pH using the ICE table and Henderson Equation?

Answer 1

Ans. Step 1: Preparation of original buffer:

Moles of NaC₂H₃O₂ = Mass / MW = 0.5 g / (82.034388 g/ mol) = 0.006095 mol

Note: Sodium acetate is CH_­3COONa or NaC₂H₃O₂ but NOT NaC₂H₃O₃

# Total volume made upto = 2 mL (acetic acid) + 12 mL (water) = 14.0 mL

It’s assumed that addition of 0.5 g NaC₂H₃O₂ does not affect the overall volume.

# Let denote-

            CH₃COOH = AH                   ; CH₃COO^- = A^-

# [AH] in final buffer solution:

Using C1V1 (original solution) = C2V2 (final buffer solution)

            Or, [AH], C2 = (3 M x 2.0 mL) / 14.0 mL = 0.4286 M

# [A^-] in final buffer solution = Moles of NaC₂H₃O₂ / Vol of buffer soln. in liters

                                                = 0.006095 mol / 0.014 L

                                                = 0.4354 M

# Step 2: Addition of NaOH to 5.0 mL original Buffer:

Total volume of solution = 5.0 mL + 0.2 mL = 5.2 mL

# [AH] = (0.4286 M x 5.0 mL) / 5.2 mL = 0.4121 M

[A^-] = (0.4354 M x 5.0 mL) / 5.2 mL = 0.4187 M

[NaOH] = (6.0 M x 0.2 mL) / 5.2 mL = 0.2308 M

[H₃O⁺] = 10^-14 / [OH^-] = 10^-14 / 0.2308 = 4.33 x 10^-14 M

# The two approaches of pH calculation are as follow-

Weak acid = AH = CH3COOH Conjugate base A CH3COC Given, [AH] in original Buffer = 0.4286 Origina!Vol of buffer = ;pKa-log Ka4.745 5.00 mL= ; 0.005 L kb = 10-14 / Ka = Ka of acid AH [NaOH] = Reaction: 5.556E-10 1.800E-05 6.000 M;0.200 mL NaOHA H20 0.0002 L #.1 Addition of 0.200 mL = NaOH Total volume of mixed solution = Mol of NaOHMolarity x Vol of soln in liters Following stoichiometry, 1 mol NaOH reacts with 1 mol AH to form 1 mol A So, moles of AH used up = 0.0012 mol | Initial moles of AH 0.4286 M x 0.005 L + 0.0002 L - 0.0052 L 6.000 M x 0.0002 M = 0.00120 mol 0.0021430 0.4354 M x 0.005 L = 0.002 1770 0.005 L = mol mol moles of A-formed= Remaining moles of AH = Initial moles-Moles consumed 0.00214 mol - Final [AH] in mixed solution = Remaining moles / Vol. of mixed soln. in liters = Final [A] in mixed solution Total moles of A-/ Vol. of mixed soln. in liters = 0.0012 mol l Initial moles of A-= 0.0012 mol = 0.00094 mol 0.00094 mol / 0.00338 mol / 0.00520 L = 0.00520 L = 0.1813 M 0.6494 M Now, # Using HH equation for base : pH-pKa + log ([A] / [AH]) . equation 1 pH = 4.745+log (0.6494 / 0.1813)= 4.745+log 3.581-4.745 + 0.554 5.299

AH Weak acid Reaction: Given: Ka of AH1.80E-05 Create an ICE table as shown in fiqure- :A Conjugate base Given, [NaOH [OH- So, [HO1014/[OH433E-14 M 0.2308 M H30 0.4121 0.4187 Initial (M) Change (M) 4.33E-14 [A] [H301 [AH! (X 0.419x (X4.33E-14 0.412X) Equilibrium (M) 0.412 -X 0.419 +X 4.33E-14 +X The expression can be simplified as quadratic expression as follow- Or, 1.80E-05 x (0.412-X)=X+0.419 X + Or, 7.42E-06-1.800E-05X=X" + 0.419 X + Or, X2+ Solving the quadratic equation, we get following two roots- 1.81E-14 Ka A] (Hx+a)X+ b) Or, Ka (c - X)X(a b)X ab Or, X. (a + b + Ka)X + (ab-Kac)-0 1.81 E-14 0.418718 X 7.418E-06 0 [AH] (c-X) X1 1.7715 Since X can't be negative, reject X2. Therefore, X 1.7715E-05 Now, [Hg at equilibrium4.33E-14 X4.33E-14 .77E-051.7715E-05 M Now, pHlog H30log 1.7715E-05 4.752 E-05 X2- -0418736