1. Complete the following chart: Unit 7, Lesson 08: The ph of Salt Solutions, Answers on NH 4 PO 3 3- Parent Acid or Base s the parent strong or weak? Will this ion hydrolyze? f the ion will hydrolyze (react with water), write the ionization reaction. f not, write no reaction or NR. ph of ion in solution (acidic, basic or neutral) NH 3 weak yes NH 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) acidic H 3 PO 3 weak yes PO 3 3- (aq) + H 2 O (l) HPO 3 2- (aq) + OH (aq) basic Na NaOH strong no NR neutral N 2 H 5 HCO 3 SO 4 2- N 2 H 4 weak yes N 2 H 5 (aq) + H 2 O (l) N 2 H 4 (aq) + H 3 O + (aq) acidic H 2 CO 3 weak yes HCO 3 (aq) + H 2 O (l) H 2 CO 3(aq) + OH (aq) basic H 2 SO 4 strong no NR neutral Ca 2+ Ca(OH) 2 strong no NR neutral NO 3 HNO 3 strong no NR neutral Notes and explanations: positive ions from weak bases are strong enough acids to hydrolyze. They will donate a proton to water and form the H 3 O + ion. Because this reaction forms the H 3 O + ion, it is a Ka reaction negative ions from weak acids are strong enough bases to hydrolyze. They will take a proton from water, leaving the OH- ion behind. Because this reaction forms the OH- ion, it is a Kb reaction 2. For each of the following salts: identify its parent acid and parent base and indicate if they are strong or weak for the ions which will react with water (hydrolyze) in solution, write the hydrolysis reaction state whether the salt solution is acidic, basic or neutral a) sodium nitrate: NaNO 3 (aq) parent base is NaOH, a strong base no hydrolysis parent acid is HNO 3, a strong acid no hydrolysis salt solution will be neutral b) ammonium chloride (NH 4 Cl) parent base is NH 3, a weak base hydrolysis: NH 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) parent acid is HCl, a strong acid no hydrolysis c) lithium oxalate (Li 2 C 2 O 4 ) parent base is LiOH, a strong base no hydrolysis 2-2- parent acid is H 2 C 2 O 4, a weak acid hydrolysis: C 2 O 4 (aq) + H 2 O (l) HC 2 O 4 (aq) + OH (aq) d) silver bromide (AgBr) parent base is AgOH, a weak base hydrolysis: Ag (aq) + H 2 O (l) AgOH (aq) + H 3 O + (aq) parent acid is HBr, a strong acid no hydrolysis
e) magnesium fluoride (MgF 2 ) parent base is Mg(OH) 2, a strong base no hydrolysis parent acid is HF, a weak acid hydrolysis: F (aq) + H 2 O (l) HF (aq) + OH (aq) f) potassium hydrogen carbonate (KHCO 3 ) parent base is KOH, a strong base no hydrolysis parent acid is H 2 CO 3, a weak acid hydrolysis: HCO 3 (aq) + H 2 O (l) H 2 CO 3 (aq) + OH (aq) g) barium hydrogen phosphate (BaHPO 4 ) parent base is Ba(OH) 2, a strong base no hydrolysis 2- parent acid is H 3 PO 3, a weak acid hydrolysis: HPO 4 (aq) + H 2 O (l) H 2 PO 4 (aq) + OH (aq) h) strontium sulfate (SrSO 4 ) parent base is Sr(OH) 2, a strong base no hydrolysis parent acid is H 2 SO 4, a strong acid no hydrolysis salt solution will be neutral i) calcium oxide (CaO) this is a strong base (remember?) so the solution will be basic j) (N 2 H 5 )ClO 4 parent base is N 2 H 4, a weak base hydrolysis: N 2 H 5 (aq) + H 2 O (l) N 2 H 4 (aq) + H 3 O + (aq) parent acid is HClO 4, a strong acid no hydrolysis
3. Calculate the ph of the following solutions: a) 1.60 M solution of NH 4 Cl parent base is NH 3, a weak base hydrolysis: NH 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) parent acid is HCl, a strong acid no hydrolysis This reaction shows the formation of an acidic solution, so we need to use the Ka for NH 4 +. NH 4 + is the conjugate acid of NH 3, so calculate Ka as follows: Ka = Kw / Kb = 1.0 x 10-14 / 1.8 x 10-5 = 5.56 x 10-10 NH 4 (aq) NH 3 (aq) H 3 O + (aq) 1.60 M (assume complete dissociation) E 1.60 - x x x Ka = [NH 3 ] [H 3 O + ] [NH 4 ] 5.56 x 10-10 = x 2 1.60 x = 2.98 x 10-5 mol/l 1.60 is much greater than 500, so ignore -x 5.56 x 10-10 so [H 3 O + ] at eq m = 2.98 x 10-5 mol/l ph = - log [2.98 x 10-5 ] = 4.53 (2 decimals, because 2 sd) (this makes sense- we had predicted an acidic solution)
b) 0.0155 M solution of NaHCO 3 parent base is NaOH, a strong base no hydrolysis parent acid is H 2 CO 3, a weak acid hydrolysis: HCO 3 (aq) + H 2 O (l) H 2 CO 3 (aq) + OH (aq) This reaction shows the formation of a basic solution, so we need to use the Kb for HCO 3 -. HCO 3 is the conjugate base of H 2 CO 3, so calculate Kb as follows: Kb = Kw / Ka = 1.0 x 10-14 / 4.5 x 10-7 = 2.2 x 10-8 Use an CE table to calculate the concentration of the OH- in the solution: HCO 3 (aq) H 2 CO 3 (aq) OH (aq) 0.0155 M (assume complete dissociation) E 0.0155 - x x x Kb = [H 2 CO 3 ] [OH ] [HCO 3 ] 2.2 x 10-8 = x 2 0.0155 x = 1.8 x 10-5 mol/l 0.0155 is much greater than 500, so ignore -x 2.2 x 10-8 so [OH-] at eq m = 1.8 x 10-5 mol/l poh = - log [1.8 x 10-5 ] = 4.73 (2 decimals, because 2 sd) ph = 14 poh = 14 4.73 = 9.27 (this makes sense- we had predicted a basic solution)
c) 0.750 M solution of KNO 2 parent base is KOH, a strong base no hydrolysis parent acid is HNO 2, a weak acid hydrolysis: NO 2 (aq) + H 2 O (l) HNO 2 (aq) + OH (aq) This reaction shows the formation of a basic solution, so we need to use the Kb for NO 2 -. NO 2 is the conjugate base of HNO 2, so calculate Kb as follows: Kb = Kw / Ka = 1.0 x 10-14 / 5.6 x 10-4 = 1.79 x 10-11 Use an CE table to calculate the concentration of the OH- in the solution: NO 2 (aq) HNO 2 (aq) OH (aq) 0.750 M (assume complete dissociation) E 0.750 - x x x Kb = [HNO 2 ] [OH ] [NO 2 ] 1.79 x 10-11 = x 2 0.750 x = 3.664 x 10-6 mol/l 0.750 is much greater than 500, so ignore -x 1.79 x 10-11 so [OH-] at eq m = 3.664 x 10-6 mol/l poh = - log [3.664 x 10-6 ] = 5.44 (2 decimals, because 2 sd) ph = 14 poh = 14 5.44 = 8.56 (this makes sense- we had predicted a basic solution)
d) 0.335 M solution of (N 2 H 5 )ClO 3 parent base is N 2 H 4, a weak base hydrolysis: N 2 H 5 (aq) + H 2 O (l) N 2 H 4 (aq) + H 3 O + (aq) parent acid is HClO 3, a strong acid no hydrolysis This reaction shows the formation of an acidic solution, so we need to use the Ka for N 2 H 5. N 2 H 5 is the conjugate acid of N 2 H 4, so calculate Ka as follows: Ka = Kw / Kb = 1.0 x 10-14 / 1.3 x 10-6 = 7.69 x 10-9 N 2 H 5 (aq) N 2 H 4 (aq) H 3 O + (aq) 0.335 M (assume complete dissociation) E 0.335 - x x x Ka = [N 2 H 4 ] [H 3 O + ] [N 2 H 5 ] 7.69 x 10-9 = x 2 0.335 x = 5.08 x 10-5 mol/l 0.335 is much greater than 500, so ignore -x 7.69 x 10-9 so [H 3 O + ] at eq m = 5.08 x 10-5 mol/l ph = - log [5.08 x 10-5 ] = 4.29 (2 decimals, because 2 sd) (this makes sense- we had predicted an acidic solution)