Question

Calculate the concentrations of H,C204, HC204,2042, and I species in a 0.170 M oxalic acid solution (K., - 6.5 10, K,, -6.1 * 10-S). Report your answers to 2 significant figures. [H2C,04] O M [HC,04 ]- M [C,042-] - X 10 M Enter your answer in scientific notation. [H] - M

Answer 1

Given, Molarity of H2C204 = 0.1700 M H2C204 is a weak acid, it dissociate partially into acetate and H+ in aqueous media as shown below H2C204 (aq) <-----> HC204- (aq) + H+ (aq) Initial 0.1700 0 0 Change -X +x +x Equilibrium (0.170-x) Kal value is very high , so we must use quadratic equation to find x value First dissociation constant of oxalic acid Ka=(HC204-) (H+) = (H2C204) - x) 6.50E-02 6.50E-02 x2 = = = = + x * x = ( 0.1700 x2 = ( 0.1700 . x) 6.50E-02 * ( 0.1700 - 1.11E-02 - 6.5E-02 x 6.50E-02 x - 1.105E-02 = 0 x) x2 --b for the above quadratic equation a= 1.0 b= 6.50E-02 -4ac 2a c= - 1.105E-02 x= -0.14253 on solving above equation x = 0.0775 or Now consider possitive x value to get [H+] From ICE table x = [H+] =[HC204-]= 0.07753 M The second dissciation of oxalic acid can be shown below + HC204- (aq) 0.0775 <-----> C204 2- (aq) 0 Initial Change Equilibrium H+ (aq) 0.07753 + y (0.0775 + y) - y (0.0775 - y) Given Kal = 6.50E-02 Ka2 = 6.10E-05 Weak acid dissociation constant Ka2 = ( C204 2-)(H+)/(HC204-) 6.10E-05 = (0.0775 + y) *y - (0.0775 - y) 6.10E-05 = 0.0775 *y = 0.0775 6.10E-05 = y * 1.0 SO y = 6.10E-05 since Ka is small, so we can assume (0.0775 - y) 0.0775 (0.0775 + y) - 0.0775 + 6.10E-05 = 7.76E-02 Totla [H+] concnetration = pH = log [H+] = -log [ (0.0775 + y) 7.76E-02 = ] = 0.0775 1.11 [H2C204] = (0.170-x) = 0.1700 - 0.07753 = 0.0925 M [HC204 -] = (0.0775 - y) = 0.07753 - 6.10E-05 = 7.75E-02 M [C204 2-] = y = 6.10E-05 M

Given, Molarity of H2CO3 = 0.045 M carbonic acid is a diprotic acid. The First dissociation of carbonic acid can be shown as below H2CO3(aq + H20 (1) <-----> HCO3- (aq) + H3O + (aq) Initial 0.0450 Change + x Equilibrium (0.045-x) + X Given Kal of Carbonic = 4.20E-07 Weak acid dissociation constant Kal = (HCO3-)(H30+)/(H2CO3) 4.20E-07 = x*x (0.045-x) 4.20E-07 = x^2 = 0.045 x^2 = 4.20E-07 * 0.045 sox ^2 = 1.89E-08 x = 0.00014 assume x is small then (0.045-x) will be 0.045 5% rule x/ initial con x 100 = 0.00014 - 0.045 x 100 = 0.306 % < 5% H3O+ (aq) 0.00014 Second dissociation of carbonic acid Ka2 is HCO3- (aq) + H20 (1) <-----> CO3 2- (aq) + Initial 0.00014 Change - y + y Equilibrium (0.00019-y) + y (0.00019+y) | Given Ka2= 4.80E-11 Weak acid dissociation constant Ka2= (CO3 2-)(H3O+)/(HCO3-) 4.80E-11 = y (0.00019+y) 4.80E-11 = y * 0.00014 4.80E-11 = y * 1.00 so 4.80E-11 = y = 4.800E-11 (0.00019-y) 0.00014 | 4.80E-11 = Total [H3O+] = pH = log [H3O+] + 3.86 1.37E-04 0.00019+y) = = -log[ 1.37E-04 0.00014 ] = [H2CO3] = (0.045-x) = 0.045 - 0.00014 0.045 M [HCO3 -] = 0.00325 -y= 0.00014 - 4.80E-11 = 1.37E-04 M [CO3 2-) = y = 4.80E-11 M