Question

i need help with A until D

A 106.2 mL sample of 0.115 M methylamine (CH3NH2; K = 3.7 x 10-4) is titrated with 0.240 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. You may want to reference (Pages 682 - 686) Section 16.9 while completing this problem.

Answer 1

ml 0.1062 L Molarity of CH3NH2 = 0.115 M Volume of CH3NH2 106.20 ml = We know, Moles = Molarity (M) x Volume (V in L) Moles of CH3NH2 = 0.115 mol/L * Molarity of HNO3 0.1062 L= 0.0122 mol 0.240 M Initial pH of solution or at 0.0 ml Base addition CH3NH2 is a weak base, in aq. Solution it partiallly dissociate and produce below i CH3NH2 (aq) + H2O <-------> CH3NH3+ (aq) + OH- (aq) Initial 0.1150 -- 0.000 0.000 Change + X + X Equil (0.115-X) Weak Base dissociation constant Kb 3.700E-04 (CH3NH3+) * (OH-) Weak dissociation kb (CH3NH2) 3.700E-04 (x) (X) (0.115-X) Assume X is small, Then X^2 3.700E-04 0.1150 X^2 3.700E-04 0.115 4.26E-05 6.52E-03 But from ICE table X 6.52E-03 = [OH-] The relation b/w pOH and (OH-) is pOH = -log(OH-] = -log [ 6.52E-03] = 2.1856 pH = 14 - pOH = 14 - 2.1856 11.81 so pH of solution = 11.8 0.1062 L pH after addition of 25.40 ml of Acid Molarity of CH3NH2 = 0.115 M Volume of CH3NH2 106.20 ml = We know, Moles = Molarity (M) Volume (V in L) Moles of CH3NH2 0.115 mol/L * Molarity of HNO3 0.240 M Volume of HNO3 25.400 ml = Moles of [H+] 0.240 mol/L * 0.1062 L= 0.01221 mol 0.02540 L 0.02540 L= 0.00610 mol Kb of CH3NH2 = 3.70E-04 CH3NH2 pkb = -log kb = -log [ 3.70E-04 ] = 3.432 pka = 14 - pka = 10.568 When H+ is added to Weak Base, it reacts with weak base and produce Conjugate acid We know both CH3NH2 & CH3NH3+ act as a buffer system. The ICE table is shown below. Net ionic equation after addition of H+ to the Buffer CH3NH2 (aq) + H+ (aq) <-------> CH3NH3+ (aq) Initial 0.01221 0.00610 0.000 Change -0.00610 -0.00610 + 0.00610 0.00612 0.000 0.00610 Henderson equation for buffer solution is r CH3NH2 рн pka + log L CH3NH3+ ] Equil Here the Base is CH3NH2 and Acid is CH3NH3+ pH = 10.568 + 0.0061 1 log L 0.0061 log[ 1.0034 ] 0.001 = 10.570 + 10.568 10.568 + pH of solution 10.6

Part C :

Molarity of CH3NH2 0.115 M Volume of CH3NH2 106.20 ml = 0.1062 L We know, Moles = Molarity (M) x Volume (V in L) Moles of CH3NH2 0.115 mol/L * 0.1062 L= 0.01221 mol Molarity of HNO3 0.240 M Volume of HNO3 50.900 ml = 0.05090 L Moles of [H+] 0.240 mol/L * 0.05090 L= 0.01222 mol Total volume of solution = V of Acid + V of Base = 0.1062 + 0.0509 = 0.1571 L Net ionic equation after addition of H+ to CH3NH2 CH3NH2 (aq) + H+ (aq) <-------> CH3NH3+ (aq) Initial 0.0122 0.0122 0.000 Change - 0.0122 - 0.0122 + 0.0122 Equil 0.0000 0.000 0.0122 From the above ICE table shows the CH3NH2 moles = 0.0122 mol of CH3NH2 Concetration of CH3NH3+ = 0.0122 mol : 0.1571 L = 0.078 M CH3NH3+ is a weak acid, it dissociate partially into CH3NH2 and H3O+ in aqueous media <-------> + CH3NH3+ (aq) 0.078 Initial Change Equilib CH3NH2(aq) 0.00 + x H+ (aq) 0.00 + x -X (0.078-X) * 100 * 100 % Kb of CH3NH2 = 3.70E-04 Ka = kw/Kb = 1.00E-14 3.70E-04 = 2.703E-11 Weak acid dissociation constant Ka 5.55E-10 (CH3NH2) (H+) 5% Rule Weak acid dissociation ka (CH3NH3+) (X)(X) Initial C 5.55E-10 (0.078-X) 6.57E-06 Assume X is small, Then 0.078 X^2 = 5.55E-10 0.078 | 0.008 X^2 = 5.550E-10 * 0.078 4.316E-11 6.57E-06 But from ICE table 6.569E-06 = [H+] or [H3O+] The relation between pH and H3O+ is, pH = -log[H3O+] pH = - log [ -log ! 6.569E-06 ] 5.182 so pH of solution 5.18 0.008 < 5%

Part D :

0.1062 L Molarity of CH3NH2 0.115 M Volume of CH3NH2 106.20 ml = We know, Moles = Molarity (M) x Volume (V in L) Moles of CH3NH2 0.115 mol/L * Molarity of HNO3 0.240 M Volume of HNO3 76.300 ml = Moles of (H+) 0.240 mol/L * 0.1062 L= 0.01221 mol 0.07630 L 0.07630 L= 0.01831 mol <-------> Net ionic equation after addition of H+ to CH3NH2 CH3NH2 (aq) + H+ (aq) Initial 0.0122 0.0183 Change -0.0122 -0.0122 Equil 0.0000 0.00610 CH3NH3+ (aq) 0.000 + 0.0122 0.0122 Above ICE table shows the CH3NH2 & (H+) moles. H+ is stonger base than CH3NH2. SO CH3NH2 moles can be neglected Total volume of solution = V Acid + V of Base = 0.1062 Concetration of [H+] = 0.00610 mol = but pH = -log[H+] = -log[ 0.0334 ] so pH of solution 1.48 + 0.1825 0.0763 L = = 0.1825 0.0334 L M