Exam #3 study guide (10-18-2022)

Name: Student ID Number: H 2 SO 4 + 2NH 3 ❑ ↔ SO 4 2- + 2NH 4 + (aq) You can clearly see that 1 mole of H 2 SO 4 reacts with 2 moles of NH 3 (base) to produce 2 moles of NH 4 + (conjugate acid). 0.00103 moles H 2 SO 4 1 x 2 mole NH 3 1 mole H 2 SO 4 = 0.00206 molesof NH 3 equivalent ¿ 0.00103 moles H 2 SO 4 Your total solution volume is 53.7 + 5 = 58.7 mL = 0.0587 L. Now calculate your new molarity for NH 3 and NH 4 + . NH 3 adjusted moles = 0.0379122 − 0.00206 = 0.0358522 NH 4 + ¿ adjusted moles = 0.0 + 0.00206 = 0.00206 ¿ [ NH 3 ] = 0.0358522 moles 0.0587 L = 0.61077 M ¿ NH 3 + H 2 O ❑ ↔ NH 4 + + OH - By subtracting the equivalent moles of H 2 SO 4 from NH 3 and adding it to NH 4 + we have accounted for all the H 2 SO 4 added as titrant so now we can focus on NH 3 reaction with water (which is amphiprotic). Now create your I.C.E. table. NH 3 NH 4 + OH - Initial (M) 0.61077 0.035094 0.0 Change -x +x +x Equilibrium 0.61077 – x 0.035094 + x x K b = ¿¿ 1.80 x 10 − 5 = [ 0.035094 + x ] x [ x ] [ 0.61077 − x ] 0.035094 1.80 x 10 − 5 = 1949.65 ( yes wecanapproximate ) 1.80 x 10 − 5 = [ 0.035094 ] x [ x ] [ 0.61077 ] x = ( 1.80 x 10 − 5 ) x ( 0.61077 ) 0.035094 = 3.1327 x 10 − 4 M

Name: Student ID Number: 4. A 53.7 mL sample of a 0.706 M solution of NH 3 which is a weak base has a base ionization constant value of K b = 1.80 x 10 -5 at 25°C is being titrated by a 0.206 M solution of H 2 SO 4 which is a strong diprotic acid. Calculate the pH when 105.7 mL of H2SO4 has been added. excess volume = 105.7 − 92.02 = 13.68 mL = 0.01368 L 0.206 mole L x 0.01368 L 1 = 0.00281808 excess mole H 2 SO 4 Your total solution volume is 53.7 + 105.7 = 159.4 mL = 0.1594 L. [ H ¿ ¿ 2 SO 4 ] excess = 0.00281808 mole 0.1594 L = 0.017680 M ¿ The H 3 O + ion controls the pH and there is nothing left for H 2 SO 4 to react with except water which is amphiprotic. H 2 SO 4 (aq) + 2H 2 O (aq) ↔ SO 4 2- (aq) + 2H 3 O + (aq) Notice that for every 1 mole of H 2 SO 4 will react with 2 mole of H 2 O to form 2 moles of H 3 O + which is what dictates the pH. 0.00281808 mole H 2 SO 4 1 x 2 mole H 3 O + ¿ 1 mole H 2 SO 4 = 0.00563616 mole H 3 O + ¿¿ ¿ ¿ ¿ Whether you multiply the moles of H 2 SO 4 by 2 and then divide by the total solution volume or just multiply the molarity of H 2 SO 4 you get around 0.03536 M for H 3 O + . Now just simply take the negative logarithm base 10 of the [H 3 O + ] and you have the pH.