Homework #6 Chapter 7 Homework Acids and Bases

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1 Homework #6 Chapter 7 Homework Acids and Bases 20. a) 2H 2O(l) H 3O (aq) OH (aq) K [H 3 O ][OH ] Or H 2O(l) H (aq) OH (aq) K [H ][OH ] b) HCN(aq) H 2O(l) H 3O (aq) CN (aq) K [H 3O ][CN ] [HCN] Or HCN(aq) H (aq) CN (aq) K [H ][CN ] [HCN] c) NH 3(aq) H 2O(l) NH 4 (aq) OH (aq) K [NH 4 ][OH ] [NH 3 ] For this question you could have picked any weak acid or base or just used a general weak acid (HA) and weak base (B) 24. a) HClO 4 strong acid b) HOCl weak acid c) H 2SO 4 strong acid (1 st deprotonation only) d) H 2SO 3 weak acid 27. The larger the K a the stronger the acid (table 7.2). HClO 4 > HClO 2 > NH 4 > H 2O Strongest acid Weakest acid Note: HClO 4 is a strong acid, therefore, will have the highest K a value. H 2O is the conjugate acid of OH. OH is a strong base. Strong bases have the weakest conjugate acids. (H 2O(l) OH (aq) OH (aq) H 2O(l)) 28. The larger the K a the stronger the acid (table 7.2), the stronger the acid the weaker the conjugate base, therefore, the strongest base will have the conjugate acid with the smallest K a. NH 3 > ClO 2 > H 2O > ClO 4 Note: HClO 4 is a strong acid, therefore, will have the highest K a value. Acids are classified as strong acids if they have a K a greater than 1. H 3O is one of the weakest strong acids. Therefore, H 3O (H 2O conjugate base) is one of the weakest strong acid and H 2O will be slightly more basic than ClO 4. 1

2 29. The larger the K a the stronger the acid. a) HCl strong acid and H 2O is a weak acid K w K a ( ) b) HNO 2 The K a of HNO 2 ( ) > K w K a ( ) c) HCN The K a HCN ( ) > K a of HOC 6H 5 ( ) 30. The higher the K a the weaker the conjugate base because K b K w K a a) Cl conjugate base of a strong acid, HCl H 2O conjugate base of one of the weakest strong acids H 3O H 2O stronger base b) H 2O conjugate base of one of the weakest strong acids H 3O NO 2 conjugate base of a weak acid NO 2 stronger base c) CN conjugate base of the weak acid HCN (K a ) OC 6H 5 conjugate base of the weak acid HOC 6H 5 (K a ) OC 6H 5 stronger base 31. a ) CH 3CO 2H acid H 2O base CH 3CO 2 conjugate base H 3O conjugate acid Therefore, H 2O and CH 3CO 2 are competing for protons. b) CH 3CO 2 is the stronger base (H 3O is a strong acid, therefore, forms a weaker conjugate acid than CH 3CO 2H) c) CH 3CO 2 is a weak base because it is the conjugate base of a weak acid. Conjugate bases of weak acid are also weak, therefore, the do no fully dissociate in solution. CH 3CO 2 H 2O(aq) CH 3CO 2H(aq) OH (aq) 33. a) weak acid b) strong acid c) weak base d) strong base e) weak base f) weak acid g) weak acid h) strong base i) strong acid (1 st deprotonation) 40. The solution is acidic if [H ]>[OH ], the solution is basic if [H ]<[OH ], and the solution is neutral if [H ][OH ]. The relationship between [H ] and [OH ] is: K w [H ][OH ] a) ( )[OH ] [OH ] M [H ] [OH ] neutral solution b) ( )[OH ] [OH ] 12 M [H ] < [OH ] basic solution c) (12)[OH ] [OH ] M [H ] > [OH ] acidic solution d) ( )[OH ] [OH ] M [H ] > [OH ] acidic solution 2

3 41. If the ph < 7 the solution is acidic, if the ph > 7 the solution is basic, and if the ph 7 the solution is neutral. If the poh < 7 the solution is basic, if the poh > 7 the solution is acidic, and if the poh 7 the solution is neutral. The relationship between ph and [H ] is: ph log[h ] The relationship between [H ] and [OH ] is: K w [H ][OH ] a) Basic [H ] 10 ph M ( )[OH ] [OH ] M b) Basic [H ] 10 ph M ( )[OH ] [OH ] 20 M c) Acidic [H ] 10 ph M (10)[OH ] [OH ] M d) Acidic [H ] 10 ph M ( )[OH ] [OH ] M e) Basic [OH ] 10 poh M [H ] ( ) [H ] M f) Acidic [OH ] 10 poh M [H ] ( ) [H ] M 47. a) HNO 2 is a weak acid The major species in solution are HNO 2 The following equilibrium will occur in solution HNO 2(aq) NO 2 (aq) H (aq) K a (Table 7.2) Calculate the concentration of H ions at equilibrium HNO 2 NO 2 H Initial M 0 0 Change x x x Equilibrium 0.250x X X K a [NO 2 ][H ] xx [HNO 2 ] (0.250 x) Since K a is small assume x xx (0.250) x M 3

4 Check assumption (5% rule) Original Concentration % 100% 4% Good Concentration of H [H ] x M ph log[h ] log(0.010) 2.00 b) HC 2H 3O 2 is a weak acid The major species in solution HC 2H 3O 2 The following equilibrium will occur in solution HC 2H 3O 2(aq) C 2H 3O 2 (aq) H (aq) Calculate the concentration of H ions at equilibrium HC 2H 3O 2 C 2H 3O 2 H Initial M 0 0 Change x x x Equilibrium 0.250x x x K a [C 2H 3 O 2 ][H ] xx [HC 2 H 3 O 2 ] (0.250 x) Since K a is small assume 0.250x xx (0.250) x Check assumption (5% rule) K a (Table7.2) % 100% 0.84% Good Original Concentration 0.25 Concentration of H [H ] x M ph log[h ] log(0.0021) a) HOC 6H 5 is a weak acid therefore The major species in solution are HOC 6H 5 The following equilibrium will occur in solution HOC 6H 5(aq) OC 6H 5 (aq) H (aq) K a Calculate the concentrations of H at equilirium HOC 6H 5 OC 6H 5 H Initial M 0 0 Change x x x Equilibrium x x x K a [OC 6H 5 ][H ] xx [HOC 6 H 5 ] (0.250 x) Since K a is small assume 0.250x xx (0.250) x Check assumption (5% rule) Original Concentration % 100% % Good

5 Concentration H [H ] x M ph log[h ] log( ) 5.20 b) HCN is a weak acid The major species in solution are HCN The following equilibrium will occur in solution HCN(aq) CN (aq) H (aq) K a (Table 7.2) Calculate the concentration of H ion at equilibrium HCN CN H Initial M 0 0 Change x x x Equilibrium 0.250x x x K a [CN ][H ] xx [HCN] (0.250 x) Since K a is small assume 0.250x xx (0.250) x Check assumption (5% rule) % 100% % Good Original Concentration 0.25 Concentration H [H ] x M ph log[h ] log( ) HF is a weak acid therefore, the reactions that are occurring in solution are: HF(aq) F (aq) H (aq) K a (Table 7.2) Calcualte the concentrations of HF, F, and H at equilibrium (you will also need to calculate the consentration of OH which will be in the soltion from the following equilibrium H 2O(l) OH (aq) H (aq)) HF F H Initial M 0 0 Change x x x Equilibrium 0.020x x x K a [F ][H ] xx [HF] (0.250 x) x x 2 x x x b ± b2 4ac 2a ± ( ) and x is M because concentration cannot be negative. Find the final concentrations [HF] 0.20 x M [F ] [H ] x M 5

6 Calculate the concentration of OH from water equilibrium K w [H ][OH ] (0.0035)[OH ] [OH ] M ph log[h ] log(0.0035) a) HA is a weak acid. There are more HA molecules (green and white atoms together) than H ions (single white atom) and A ions (single green atoms). This shows that the acid did not completely dissociate making it a weak acid. b) Calculate the percentage dissociated In the picture there are 9 particles that have note dissociated and 1 particle that has particles dissociated 100% 1 100% 10% total particles 1 9 Calculate the equilibrium constant The reaction of interest is HA(aq) H (aq) A (aq) HA H A Initial 0.20 M 0 0 Change x x x Equilibrium x x 1 N a V N a V x 1 N a V Note from the picture at equilibrium the concentration of HA is 9 Na 9 Na is V N a V Avogadro s number and is used to convert the 9 particles to number of moles of HA. The concentration of H and A are Solve for V 0.20 x 9 N A V N a V 9 N a V V 10 N a N a Solve for K K [H ][A ] [HA] 1 N a ( 50 1 Na ) N a ( 50 Na ) 9 N a ( 50 N a ) 1 N a V 1 (50)(9) HC 2H 2ClO 2 is a weak acid, therefore, the equilibrium established is: HC 2H 2ClO 2(aq) C 2H 2ClO 2 (aq) H (aq) K a Find the concentration of H at equilirbium HC 2H 2ClO 2 C 2H 2ClO 2 H Initial 0.10 M 0 0 Change x x x Equilibrium 0.10x x x 6

7 K a [C 2H 2 ClO 2 ][H ] xx [HC 2 H 2 ClO 2 ] (0.10 x) x x 2 x x x b ± b2 4ac 2a ± ( ) and x is M because concentration cannot be negative. [H ] x M ph log[h ] log(0.011) HIO 3 is a weak acid, therefore, the equilibrium established is: HIO 3(aq) IO 3 (aq) H (aq) K a 0.17 Find the concentration of H at equilibrium HIO 3 IO 3 H Initial M 0 0 Change x X x Equilibrium 0.010x x x K a [IO 3 ][H ] [HIO 3 ] x x 2 x x x b ± b2 4ac xx (0.10 x) ± (0.17) and a x is M because concentration cannot be negative. [H ] x M ph log[h ] log(0.0094) g C 6 H 5 CO 2 H ( 1 mol C 6H 5 CO2H g C6H 5 CO2H 6H 5 CO 2 H M n mol M V 1.0 L C 6H 5CO 2H is a weak acid, therefore, the equilibrium established is: C 6H 5CO 2H(aq) C 6H 5CO 2 (aq) H (aq) K a Calcualte the equilirium concnetrations C 6H 5CO 2H C 6H 5CO 2 H Initial M 0 0 Change x x x Equilibrium x X X K a [C 6H 5 CO 2 ][H ] xx [HC 6 H 5 CO 2 ] ( x) x x 2 x x

8 x b ± b2 4ac 2a ± ( ) and x is M because concentration cannot be negative. Find the final concentrations [HC 6 H 5 CO 2 ] x M [C 6 H 5 CO 2 ] [H ] x M Calculate the concentration of OH from water equilibrium K w [H ][OH ] ( M)[OH ] [OH ] M ph log[h ] log( ) C 6H 5CO 2H is a solid before it is put into water. Therefore, two equilibriums are established. C 6H 5CO 2H(s) C 6H 5CO 2H(aq) C 6H 5CO 2H(aq) C 6H 5CO 2 (aq) H (aq) K a They want to know how much C 6H 5CO 2H will dissolve. This will be the amount of C 6H 5CO 2H that starts in the second equilibrium which we will denote as s. C 6H 5CO 2H C 6H 5CO 2 H Initial S 0 0 Change x x x Equilibrium sx x x The ph of the solution will give us the equilibrium concentration of the H ions ph log[h ] [H ] 10 ph M C 6H 5CO 2H C 6H 5CO 2 H Initial s 0 0 Change Equilibrium s K a [C 6H 5 CO 2 ][H ] (0.0016)(0.0016) [C 6 H 5 CO 2 H] (S ) S M Convert to grams per 100 ml mol S ( ) ( 1 L ) 1 L 1000 ml ( g C 6H 5 CO 2 H ) g 1 mol C 6 H 5 CO 2 H ml Therefore the water solubility per 100 ml is 0.51 g per 100 ml 61. a) The first solution contains 0.10 M HCl (strong acid) and 0.10 M HOCl (weak acid). Since there is both a strong and weak acid with equal concentrations of each the number of H ions from the weak acid will be minimal, therefore, the ph will be completely from the strong acid. ph log[h ] log(0.10) 1.00 This can be proven using an ICE table HOCl(aq) H (aq) OCl (aq) K a HOCl OCl H 8

9 Initial 0.10 M Change x x x Equilibrium 0.10x x 0.10x K a [OCl ][H ] x(0.10 x) [HOCl] (0.10 x) Since K a is small assume 0.10x 0.10 and 0.10x0.10 x(0.10) (0.10) x Check assumption (5% rule) Original Concentration % 100% % 0.10 Good Concentration H [H ] 0.10 x M ph log[h ] log(0.10) 1.00 b) The second solution contains M HNO 3 (strong acid) and 0.50 M HC 2H 3O 2 (weak acid). Since there is both a strong and weak acid and even though the concentration of the weak acid is 10x more than the concentration of the strong acid the number of H ions from the weak acid will be minimal, therefore, the ph will be completely from the strong acid. ph log[h ] log This can be proven using an ICE table HC 2H 3O 2(aq) H (aq) C 2H 3O 2 (aq) K a HC 2H 3O 2 C 2H 3O 2 H Initial 0.50 M Change x x x Equilibrium 0.50x x 0.050x K a [C 2H 3 O 2 ][H ] x(0.050 x) [HC 2 H 3 O 2 ] (0.50 x) Since K a is small assume 0.50x 0.50 and 0.050x0.050 x(0.050) (0.50) x Check assumption (5% rule) Note:0.050<0.50 therefore is the assumption works for 0.050x then it will work form 0.50x % 100% 0.36% Good Original Concentration Concentration H [H ] x M ph log[h ] log(0.50)

10 65. HOCN is a weak acid, therefore, the following equilibrium is established: HOCN(aq) OCN (aq) H (aq) HOCN OCN H Initial M 0 0 Change x x x Equilibrium x x x The ph of the solution at equilibrium will allow us to calculate the concentration of H ions at equilibrium x ph log[h ] [H ] 10 ph M Updating the ICE table gives HOCN OCN H Initial M 0 0 Change Equilibrium K a [OCN ][H ] [HOCN] (0.0017)(0.0017) (0.0083) The larger the K b the stronger the base. Therefore the strongest base will have the largest K b. NH 3 > C 5H 5N > H 2O > NO 3 Strongest base Weakest base Note: HNO 3 is a strong acid. The stronger the acid the weaker the conjugate base. Therefore, NO 3 is a very weak base. Note: H 2O is the conjugate base of (H 3O ). H 3O is one of the weakest strong acids. Therefore, H 2O is only slightly more basic than NO The smaller the K b the stronger the conjugate acid. HNO 3 > C 5H 5NH > NH 4 > H 2O Strongest acid Weakest acid Note: HNO 3 is a strong acid. Note: C 5H 5N and NH 3 are both weak bases, therefore, will have weak conjugate acids. The K b of NH 3 > k b of C 5H 5N resulting in C 5H 5NH being a stronger acid than NH 4. Note: H 2O is the conjugate acid of OH. OH is a strong base. Strong bases have the weakest conjugate acids. (H 2O(l) OH (aq) OH (aq) H 2O(l)) 74. a) Ca(OH) 2 is a strong base (fully dissociates) [OH ] M Ca(OH) 2 ( 2 mol OH 1 mol Ca(OH) 2 poh log[oh ] log( ) 3.10 ph poh ph poh b) KOH is a strong base (fully dissociates) 25 g KOH ( [OH ] n V 1 mol KOH mol OH ) ( g KOH 1 mol KOH 0.45 mol 1.00 L 0.45 M ) 0.45 mol OH ) M OH 10

11 poh log[oh ] log(0.45) 0.35 ph poh ph poh c) NaOH is a strong base (fully dissociates) 1 mol NaOH 1 mol OH g NaOH ( ) ( ) mol OH g NaOH 1 mol NaOH [OH ] n mol M V L poh log[oh ] log(3.750) ph poh ph poh Ba(OH) 2 is a strong base (fully dissociates) Calculate the concentration of OH ph poh poh ph poh log[oh ] [OH ] 10 poh Calculate the concentration of Ba(OH) M ( 1 mol Ba(OH) 2 1 mol OH ) M Ba(OH) The presence of nitrogen most often results in a basic molecule. Nitrogen has an unpaired electron, therefore, bonds easily to H ions The larger the ph the more basic a solution is a) HI strong acid HF weak acid NaF (weak base) NaF(s) Na (aq) F (aq) Na H 2O NaOH H (NaOH is a strong base) Na is neutral F H 2O HF OH F is basic NaI (neutral) NaI(s) Na (aq) I (aq) Na H 2O NaOH H (NaOH is a strong base) Na is neutral I H 2O HI OH (HI is a strong acid) I is neutral HI < HF < NaI < NaF Increasing ph b) NH 4Br (weak acid) NH 4Br(s) NH 4 (aq) Br (aq) NH 4 H 2O NH 3 H 3O NH 4 is acidic Br H 2O Br is neutral HBr OH (HBr is a strong acid) 11

12 HBr strong acid KBr (neutral) KBr(s) K (aq) Br (aq) K H 2O KOH H (KOH is a strong base) K is neutral Br H 2O HBr OH (HBr is a strong acid) Br is neutral NH 3 weak base HBr < NH 4Br < KBr < NH 3 Increasing ph c) C 6N 5NH 3NO 3 (weak acid) C 6N 5NH 3NO 3(s) C 6N 5NH 3 (aq) NO 3 (aq) C 6N 5NH 3 H 2O C 6N 5NH 2 H 3O C 6N 5NH 3 is acidic NO 3 H 2O HNO 3 OH (HNO 3 is a strong acid) NO 3 is neutral NaNO 3 (neutral) NaNO 3(s) Na (aq) NO 3 (aq) Na H 2O NaOH H (NaOH is a strong base) Na is neutral NO 3 H 2O HNO 3 OH (HNO 3 is a strong acid) NO 3 is neutral NaOH strong base HOC 6H 5 weak acid (K a ) KOC 6H 5 (weak base) KOC 6H 5(s) K (aq) OC 6H 5 (aq) K H 2O KOH H (KOH is a strong base) K is neutral OC 6H 5 H 2O HOC 6H 5 OH OC 6H 5 is basic C 6H 5NH 2 weak base (K b ) HNO 3 Strong acid There are 2 weak acid C 6N 5NH 3NO 3 (C 6H 5NH 3 ) and HOC 6H 5 Compare the K a values to determine the weaker acid HOC 6H 5 K a The K a value of C 6H 5NH 3 is not given therefore, it must be calculated C 6H 5NH 2(aq) K b K a K w K b Therefore, C 6N 5NH 2NO 3 is a stronger acid than HOC 6H 5 There are 2 weak bases C 6N 5NH 2 and KOC 6H 5 (OC 6H 5 ) Compare the K b values to determine the weaker base C 6N 5NH 2 K b The K b value of OC 6H 5 is not given therefore, it must be calculated HOC 6H 5(aq) K a

13 K a K w K b Therefore, KOC 6H 5 is a stronger base than C 6N 5NH 2 HNO 3< C 6N 5NH 3NO 3 < HOC 6H 5 < NaNO 3< C 6H 5NH 2 < KOC 6H 5 < NaOH 103. See chart for problem 85. a) Sr(NO 3) 2 (salt formed form a strong acid and strong base) Sr(NO 3) 2 Sr(NO 3) 2(s) Sr 2 (aq) 2NO 3 (aq) Sr 2 2H 2O Sr(OH) 2 2H (Sr(OH) 2 is a strong base) Sr 2 is neutral NO 3 H 2O HNO 3 OH (HNO 3 is a strong acid) NO 3 is neutral Therefore, Sr(NO 3) 2 is neutral b) C 2H 5NH 3CN C 2H 5NH 3CN(s) C 2H 5NH 3 (aq) CN (aq) C 2H 5NH 3 H 2O C 2H 5NH 2 H 3O C 2H 5NH 3 is acidic CN H 2O HCN OH CN is basic Therefore, need to compare K a and K b values of reactions We are not given the K a of C 2H 5NH 3, therefore, we need to calculate it C 2H 5NH 2(aq) K b K a K w K b We are not given the K b of CN, therefore, we need to calculate it HCN(aq) K a K b K w K a Since K a < K b C 2H 5NH 3CN is basic c) C 5H 5NHF C 5H 5NHF(s) C 5H 5NH (aq) F (aq) C 5H 5NH H 2O C 5H 5N H 3O C 5H 5NH is acidic F H 2O HF OH F is basic Therefore, need to compare K a and K b values of reactions We are not given the K a of C 5H 5NH, therefore, we need to calculate it C 2H 5N(aq) K b K a K w K b We are not given the K b of F, therefore, we need to calculate it HF(aq) K a K b K w K a Since K a > K b C 5H 5NHF is acidic 13

14 d) NH 4C 2H 3O 2 NH 4C 2H 3O 2(s) NH 4 (aq) C 2H 3O 2 (aq) NH 4 H 2O NH 3 H 3O NH 4 is acidic C 2H 3O 2 H 2O HC 2H 3O 2 OH C 2H 3O 2 is basic Therefore, need to compare K a and K b values of reactions NH 4 K a We are not given the K b of C 2H 3O 2 therefore we need to calculate it HC 2H 3O 2(aq) K a K b K w K a K a K b NH 4C 2H 3O 2 is neutral e) NaHCO 3 NaHCO 3(s) Na (aq) HCO 3 (aq) Na H 2O Na(OH) H (NaOH is a strong base) Na is neutral HCO 3 H 2O H 2CO 3 OH HCO 3 H 2O CO 3 2 H 3O Since H 2CO 3 is a polyprotic acid must look at K a and K b of the two possible reactions. HCO 3 K a We are not given the K b of HCO 3 therefore we need to calculate it H 2CO 3(aq) K a K b K w K a Since K a < K b HCO 3 is basic Therefore, NaHCO 3 is basic a) KCl KNO 2 KCl(s) K (aq) Cl (aq) K H 2O KOH H (KOH is a strong base) K is neutral Cl H 2O HCl OH (HCl is a strong acid) Cl is neutral Therefore, KCl is neutral Therefore, the solution will be neutral ph7.00. For neutral solutions the concentration of H and OH equals M. b) KF KF(s) K (aq) F (aq) K H 2O KOH H (KOH is a strong base) K is neutral F H 2O HF OH F is basic Therefore, KF is basic 14

15 Interested in the following reaction F (aq) H 2O(l) HF(aq) OH (aq) Problem we do not know K b HF(aq) K a K b K w Calculate the concentration of OH ions F HF OH Initial 1.0 M 0 0 Change x x x Equilibrium 1.0x x x K b [OH ][HF] xx K a [F ] 1.0 x Since K b is small assume that 1.0x 1.0 xx x Check assumption (5% rule) Original Concentration Calculate OH concentration [OH ] x M Calculate H concentration [H ][OH ] [H ] % 100% % Good ph log[h ] log( ) a) KNO 2 KNO 2(s) K (aq) NO 2 (aq) K H 2O KOH H (KOH is a strong base) K is neutral NO 2 H 2O HNO 2 OH NO 2 is basic Therefore, KNO 2 is basic Interested in the following reaction NO 2 (aq) H 2O(l) HNO 2(aq) OH (aq) Problem we don t know K b Know HNO 2(aq) K a K b K w Calculate the concentration of OH ions NO 2 HNO 2 OH Initial 0.12 M 0 0 Change x x x Equilibrium 0.12x x x K a

16 K b [HNO 2][OH ] xx [NO ] 0.12 x Since K b is small assume that 0.12x 0.12 xx x M Check assumption (5% rule) Original Concentration % 100% % Good 0.12 Calculate OH concentration [OH ] x M poh log[oh ] log( ) ph poh ph poh b) NaOCl NaOCl(s) Na (aq) OCl (aq) Na H 2O NaOH H (NaOH is a strong base) Na is neutral OCl H 2O HOCl OH OCl is basic Therefore, NaOCl is basic Interested in the following reaction OCl (aq) H 2O(l) HOCl(aq) OH (aq) Problem we don t know K b HOCl(aq) K a K b Calculate the concentration of OH ions OCl HOCl OH Initial 0.45 M 0 0 Change x x x Equilibrium 0.45x X x K b [HOCl][OH ] xx [OCl ] 0.45 x Since K b is small assume that 0.45x 0.45 xx x M Check assumption (5% rule) Original Concentration % 100% 0.080% Good 0.45 Calculate OH concentration [OH ] x M poh log[oh ] log( ) 3.44 ph poh ph poh

17 c) NH 4OCl 4 (a salt from a weak base and a strong acid) NH 4OCl 4(s) NH 4 (aq) OCl 4 (aq) NH 4 H 2O NH 3 H 3O NH 4 is acidic OCl 4 H 2O HOCl 4 OH (HOCl 4 strong acid) OCl 4 is neutral Therefore, NH 4OCl 4 is acidic Interested in the following reaction NH 4 (aq) NH 3(aq) H (aq) K a Calculate the concentration of H ions NH 4 NH 3 H Initial 0.40 M 0 0 Change x x x Equilibrium 0.40x x X K a [NH 3][H ] xx [NH ] 0.40 x Since K b is small assume that 0.40x 0.40 xx x M Check assumption (5% rule) Original Concentration % 100% % Good 0.40 Calculate H concentration [H ] x ph log[h ] log( ) Because the solution has a ph of 4.6 we know that the solution is acidic, therefore, NaOH, NH 3, and NaCN cannot be the solute. Because the bulb is bright it means that there are a lot of ions in solution. Therefore, a strong electrolyte is needed. Only HCl, NaOH, NH 4Cl, NaCN are strong electrolytes. NaOH and NaCN were already eliminated because they were basic. HCl could light the bulb bright because it is a strong acid but the ph of a 1.0 M solution of HCl is ph log[h ] log(1.0) 0 Confirm the solution is made from NH 4Cl by compare actual to calculated ph values Determine the K a of the weak acid. HA(aq) A (aq) H (aq) K a unknown HA A H Initial 1.0 M 0 0 Change x x x Equilibrium 1.0x x x 17

18 K a [A ][H ] xx [HA] (1.0 x) Determine x x [H ] ph log[h ] 4.6 log[h ] [H ] K a [A ][H ] ( )( ) [HA] This K a of NH 4 is which is close to the calculated value. Therefore the solution must contain NH 4Cl 18