Question

2. (a) Determine the pH of 0.10 M lactic acid (HC3HSO3) (b) What is the pH of a buffer that is 0.12 M. lactic acid and 0.10 M sodium lactate (NaC3H5O3)? For lactic acid. Ka = 1.4 x 10-4. (10)

Answer 1

Given, Molarity of HC3H503= 0.100 mol/L it acid is a weak acid, it dissociate partially into lactate and H3O+ in aqueous media as shown below HC3H503 (aq + H20 (1) <----->C3H503- (aq) + H3O+ (aq) Initial 0.100 0 Change - X +x Equilibrium 0.100-x Here Acid is HC3H503 and H3O+ and base is C3H503- Given Ka= 1.40E-04 Weak acid dissociation constant Ka= (C3H503-)(H3O+)/(HC3H503) 1.40E-04 = X * X ( 0.100 - x) Assune x is small then 0.100 1.40E-04 = x + x = 0.100 will be 0.100 x^2 = 1.40E-04 * 0.100 5% rule: SO X^2 = 1.40E-05 (x = initial conc)x 100 0.00374 x = 0.100 x 100 but from ICE table x = [H3O+] so [H3O+] = 0.00374 3.742 % but pH = -log [H3O+] Which is <5% = -log[ 0.003742 ] = 2.427 so pH of Acetic Acid solution = 2.42 - x

mol We know, Moles = Molarity (M) x Volume (V in L) Molarity of HC3H503= 0.120 M So Moles HC3H503 = 0.120 * 1.00 mol Molarity of C3H503- = 0.100 M = 0.120 mol Volume of buffer V= 1000 mL moles C3H503-= 0.100 * 1.00 = 1.00 L = 0.100 mol note: 1ml = 10^-3 L Given Ka= 1.40E-04 we know pKa = -log (Ka) = -log[ 1.40E-04 ] = 3.854 Henderson equation for buffer solution is pH=pKa + log [Base/Acid] here the Base is C3H503- and acid is HC3H503 pH = pka + log [C3H503- / HC3H503] = 3.854 + log [ 0.1 0.12 ] = 3.854 + log [ 0.8333 ] = 3.854 + -0.079 = 3.77