Question

Tutored Practice Problem 17.3.5 COUNTS TOWAIDA CHEE Calculate the pH and concentration of species present in a polyprotic acid solution. Close Problem For a 5.96X10 M solution of H2CO3, calculate both the pH and the Co, ion concentration. H2CO3 + H20= HCO3 + H20= H20* + HCO3 K1 = 4.2*10-7 H30* +CO;2- K2 -4.810-11 pH (CO3? Check & Submit Answer Show Approach Polyprotic Acid Titration Plots The pH plot for the titration of the diprotic acid maleic acid, HOC(CH), CO,H, with a strong base is shown in Figure 17.39

Answer 1

Given, Molarity of H2CO3 = 5.96E-02 M = 0.0596 M carbonic acid is a diprotic acid. The First dissociation of carbonic acid can be shown as below H2CO3 (aq + H20 (1) <-----> C6H606- (aq) + H3O + (aq) Initial 0.0596 Change - X + x Equilibrium (0.0596-x) + X Given Kal of Carbonic = 4.20E-07 Weak acid dissociation constant Kal = (HC6H606-)(H30+)/(H2CO3) 4.20E-07 = x*x = (0.0596-x) assume x is small then 4.20E-07 = x^2 = 0.060 (0.0596-x) will be 0.060 x^2 = 4.20E-07 * 0.060 sox ^2 = 2.50E-08 5% rule x = 0.00016 x/ initial con x 100 = 0.00016 0.060 x 100 Second dissociation of carbonic acid Ka2 is = 0.265 % < 5% HCO3- (aq) + H20 (1) <-----> CO3 2- (aq) + H3O+ (aq) Initial 0.00016 0.00016 Change - y +y Equilibrium (0.00019-y) (0.00019+y) Here Acid is H2CO3 and H3O+ and base is HC6H606- Given Ka2= 4.80E-11 Weak acid dissociation constant Ka2=(CO3 2-)(H3O+)/(HCO3-) 4.80E-11 = y (0.00019+y) (0.00019-y) 4.80E-11 = y * 0.00016 0.00016 4.80E-11 = y * 1.00 so 4.80E-11 = y = 4.800E-11 + y . . 4.80E-11 = 1.58E-04 Total [H3O+] = pH = log [H3O+] 0.00019+y) = = -log[ 1.58E-04 0.00016 ] = + 3.80 [H2CO3] = (0.0596-x) = 0.060 - 0.00016 = 0.059 M [HCO3-) = 0.00325 -y= 0.00016 - 4.80E-11 1.58E-04 M [CO3 2-) = y = 4.80E-11 M