Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section

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1 Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section 1. If 1.0 liter solution has 5.6mol HCl, 5.mol NaOH and 0.0mol NaA is added together what is the ph when the resulting solution reaches equilibrium?(k a of HA is )(Answer:4.9) Limiting reagent reactions: a. HCl (aq) + OH - (aq) H 2 O(l) + Cl - (aq) Initial: 5.6mol 5.mol (L.R.) End: 0.mol 0 b. HCl (aq) + A - (aq) HA(aq) + Cl - (aq) Initial: 0. mol 0.mol End: 0mol 0mol 0.mol Equilibrium: HA(aq) + H 2 O(l) H O + (aq) + A - (aq) I 0.M 10-7 M 0.M Q<K C -x +x +x EQ 0.M-x 0.M x x x K a = ][ H O [ HA] ] ph= If 1.0 liter solution has 5.0mol HCl, 5.60mol NaOH and 0.0mol HA is added together what is the ph when the resulting solution reaches equilibrium?(k a of HA is ) (Answer:11.9) Limiting reagent reactions: a. HCl (aq) + OH - (aq) H 2 O(l) + Cl - (aq) Initial: 5.mol 5.6mol (L.R.) End: 0mol 0.mol b. HA(aq) + OH - (aq) A - (aq) + H 2 O(l) Initial: 0. mol 0.mol End: 0mol 0mol 0.mol Equilibrium: A - (aq) + H 2 O(l) OH - (aq) + HA(aq) K b = I 0.0M 10-7 M 0.M Q<K C -x +x +x EQ 0.M-x 0.M x x x [ HA][ OH ] ] poh=2.61 PH=11.9 1

2 . If 1.0 liter solution has 5.80mol HCl, 5.0mol NaOH and 0.0mol A - is added together what is the ph when the resulting solution reaches equilibrium?(k a of HA is )(Answer:0.70) Excess of 0.20M HCl ph=0.698= If 1.0 liter solution has 4.4mol HCl, 4.6mol NaOH and 0.50mol HA is added together what is the ph when the resulting solution reaches equilibrium?(k a of HA is )(Hint: To decide which equilibrium to use compare K a and K b ) (Answer:4.1) Write the chemical reaction(s) that will precede 100%? (Hint: limiting reagent) Limiting reagent reactions: a. HCl (aq) + OH - (aq) H 2 O(l) + Cl - (aq) Initial: 4.4mol 4.6mol (L.R.) End: 0mol 0.2mol b. HA(aq) + OH - (aq) A - (aq) + H 2 O(l) Equilibrium: Ka>Kb Initial: 0.5 mol 0.2mol (L.R.) End: 0.mol 0mol 0.2mol HA(aq) + H 2 O(l) H O + (aq) + A - (aq) K a = I 0.M 10-7 M 0.2M Q<K C -x +x +x EQ 0.M-x 0.M x x 0.2M+x 0.2M ][ H O [ HA] ] [H O + [ HA] ]= K a ] = ; ph= If 1.0 liter solution has 4.40mol HCl, 4.60mol NaOH and 0.50mol HA is added together what is the ph when the resulting solution reaches equilibrium?(k a of HA is )(Hint: To decide which equilibrium to use compare K a and K b ) (Answer:9.11) Limiting reagent reactions: a. HCl (aq) + OH - (aq) H 2 O(l) + Cl - (aq) Initial: 4.4mol 4.6mol (L.R.) End: 0mol 0.2mol b. HA(aq) + OH - (aq) A - (aq) + H 2 O(l) Initial: 0.5 mol 0.2mol (L.R.) End: 0.mol 0mol 0.2mol 2

3 Equilibrium: Ka<K b K b = 2*10-5 need to use base and water equilibrium A - (aq) + H 2 O(l) OH - (aq) + HA(aq) K b = I 0.2M 10-7 M 0.M Q<K C -x +x +x EQ 0.2M-x 0.2M x x 0.M+x 0.M [ HA][ OH ] ] [OH - ] ]= K b [ HA ] poh=4.89 ph= = How can you make a solution that best resists change in ph? 7. Buffer Solution initially both a weak acid and its conjugate base are present in the solution with comparable concentrations. 8. Which of the following could be added to an aqueous solution of a weak acid HA to convert the solution into a buffer? NaCl(s) HA(s) NaOH(s) HCl(aq) NaA(s) 9. Which of the following could be added to an aqueous solution of a weak base A - to convert the solution into a buffer? NaCl(s) HA(s) NaOH(s) HCl(aq) NaA(s) 10. Using an expression for K a derive an expression for [H O + ] [H O + [ HA ] ]= K a ] 11. How can you make a buffer solution where [H O + ]=K a.(or ph=pk a ) [HA]=[A - ] mol NaOH(s) is added to 1L of a 1.00 M solution of an acid (HA) with Ka = Calculate [H O + (aq)] of the resulting solution. Assume the volume of the solution does not change when the NaOH(s) is added. Number of moles of HA is by 0.4 and number of moles of A - by 0.4 [H O + [ HA] moles( HA) ]= K a Ka M ph=.824 ] moles( A ) 0.4

4 1. A buffer is made by adding 6.0 mol of HA and 4.0 mol of A - to enough water to make 1.0 L of solution. (K a = 2.0 * 10-5 ) Calculate the ph of this solution. [H O + [ HA] ]= K a ] HA(aq) + H 2 O(l) H O + (aq) + A - (aq) K a = I 6mol 10-7 M 4mol Q<K C -x +x +x EQ 6M x 4M [ HA ] ph = pk a log = -log(2 * 10-5 ) log(/2) ] ph=4.52=4.52 ][ H O [ HA] ] Calculate the ph of this solution after 1.0 L of 2.0 M HCl is added to the buffer solution in question 1. (Did the ph change dramatically?) #moles of HCl < #of moles of A - Number of moles of A - is by 2 moles and number of moles of HA by 2 moles [ HA ] ph = pk a log = -log(2 * 10-5 ) log(4/1) ] ph=4.10 ph does not change dramatically from After 1.0L of 4.0 M HCl is added to the buffer solution in question 1. (Hint: what is still in the solution? Is it still a buffer?) a) Will the addition of HCl increase, decrease, or not change the ph? b) What will be the limiting reagent (make a limiting reagent table) #moles of HCl = #of moles of A - Number of moles of A - is by 4 moles and number of moles of HA by 4 moles New concentrations [HA ]=10mol/2L=5 M [H O + 5 ]= [ HA] new K a = 10 mol 210 =1* 10-2 M 2l ph=2.00 change dramatically from 4.5 because we now have an excess of weak acid 4

5 16. Calculate the ph of the solution in question 1 after 5.0 mol of HCl is added. Assume the change in volume due to the addition of HCl is negligible. #moles of HCl > #of moles of A - we now have an excess of a strong acid [H O + ]=[ HCl]= 1.0M ; ph=0 a) ph change dramatically compared to the change in question 1 and 14 excess of a strong acid 17. Calculate the ph of the solution in question 1 after 1.0L of 5 M NaOH is added. ( Hint: does the final volume of the solution change?) #moles of NaOH < #of moles of HA (weak acid only) Number of moles of HA is by 5 moles and number of moles of A - by 5 moles New concentrations : [A - ]=9 mol/2l=4.5 M [HA]=1 mol/2l=0.5 M [ HA ] ph = pk a log = -log(2 * 10-5 ) log(0.5/4.5) = =5.7 ] 18. Calculate ph of the solution in question 1 after 6 moles of NaOH(aq) is added. Assume the change in volume due to the addition of NaOH(aq) is negligible. #moles of NaOH = #of moles of HA (weak acid only) Number of moles of HOA is by 6 moles and number of moles of A - by 6 moles New concentrations: [A - ]=10mol/2L=5 M [OH - ]= poh=4.2 ph= K 10mol 10 [ HA] = 5 1L 210 w new K a =7* 10-5 M 19. Buffer Solution has ph=5 and pk a =5.. What is the ratio of the weak acid concentration to its conjugate base that is needed to make a buffer of the given ph, [HA (aq)]/[a (aq)]? Propose what concentrations of acid and base you need to achieve that ratio? [H O + [ HA ] ]= K a ] pk a ph= log ( HA ( A ) ) ( HA ) =0. ; =2 ( A ) 5

6 20. You are asked to create a buffer solution by adding NaOH(aq) to ml of a M solution of weak acid with Ka = What value of c a / c b is required for the buffer solution to have ph =.? [H O + [ HA ] ]= K a ] log ( HA ( A ) ) = pk a ph=4-.=0.7 ( HA c a / c b = ( A ) ) =5 21. You are asked to create a buffer for which c a / c b = What volume, in ml, of a M solution of NaOH must be added to ml of a 0.00 M solution of weak acid whose ionization constant is K a = ? ( HA ) c a / c b = =0.05 c ( A a =0.05c b ) L of HA molha mol of HA initially present 1L x is the number of moles of NaOH added After neutralization reaction is over number of moles of HA=0.000-x and number of moles of A = x Limiting reagent reaction: HA(aq) + OH - (aq) A - (aq) + H 2 O(l) Initial: mol x mols(l.r.) End: x 0mol x mols x 2 x 0.05 x=0.0286moles 1L moles of NaOH 0.286L or 286ml of NaOH needed (00mL) 0.100molNaOH 22. One of the most important buffering systems in the biological realm is the carbonic acid (H 2 CO ) and carbonate ion (HCO - ) system that maintains the ph of blood plasma to a relatively constant value. In blood at 7 C the K a the carbonic acid is and the concentrations of the buffer components in blood plasma are 0.24 M HCO - and 0.12 M H 2 CO. Calculate the ph of blood for these conditions. pk a =-log( )=7.1 ( A ) ph = pk a + log =7.1 + log( 0.24/0.12) =7.4 ( HA ) 6

7 a. The actual concentrations of the buffer components in blood plasma are M HCO - and H 2 CO. Circle the condition for which buffering in blood is most effective. added acid added base equal response to acid or base 2. You have a 1M solution of each of the following: HF, NaCl, NaF, HClO 2, and NH 2 OH. The data in the table below will be helpful in answering the questions. (a) The solution with the highest ph is NH 2 OH (b) The solution with the lowest ph is HClO 2 HF HClO 2 NH 2 OH K a K b Para-aminobenzoic acid, H 2 NC 6 H 4 CO 2 H or PABA, has been used in sunscreens as a UV filter as early as 1948 after in vivo studies on mice showed that PABA reduced UV damage. In the 1980 s, however, animal studies suggested that PABA might increase the risk of cellular UV damage. Since then, PABA is no longer a preferred ingredient in sunscreens. Benzoic acid, C 6 H 5 CO 2 H, is a synthetic precursor to PABA and a weak acid with K a = 6.0 x a. What is the ph of a solution made from 0.40 mol of benzoic acid in 2.0 L of water at 25 o C? ph=2.46 b. NaOH is added to the solution in (a) until just enough base has been added to completely consume all of the benzoic acid. What volume (in ml) of 0.50 M NaOH solution must be added to reach this point? 800ml of NaOH c. Write the balanced, net-ionic acid-base chemical equation that will take place in solution once all of the NaOH solution, which was just enough to consume the benzoic acid, has been added in (b). d. NaOH is added to the solution in (a) until just enough base has been added so the ph of the solution was equivalent to the pk a. What volume (in ml) of 0.50 M NaOH solution must be added to reach this point? 400ml 25. Acetic acid is a weak acid with K a = How many moles of acetate ion need to be added to 1.0 L of 0.10 M acetic acid to get a buffer with ph = 4.? Assume that added acetate ion does not change the volume of the solution. HA(aq) + H 2 O(l) H O + (aq) + A - (aq) K a = [A - ]= ][ H O ] [ HA] [H O + ]= In a 1 M NaOH solution at 25 o C, for every hydronium ion how many water molecules must be present? 7

8 [OH - ]=1M [H O + ]= [ H 2O] 55 M [H 2 O]=55.5M [ H O ] 1*10 Solubility is the number of moles of the solid that dissolves in 1L of solution to form saturated solution. K eq = K sp Solubility product constant Solubility product: MX 2 (s) <=> M 2+ (aq)+ 2X - (aq) K sp = [M 2+ (aq)][ X (aq)] 2 Oxidation Reduction Reactions 1. The K sp of CaF 2 is , what is the molar solubility s for this compound in water? a. Chemical equilibrium reaction: CaF 2 (s) <=> Ca 2+ (aq)+ 2F - (aq) K sp =4x10-11 = 4s s= (4x10-11 /4) ⅓ = M b. What is [Ca 2+ ] = s and [F - ]=2s= M 2. What is the molar solubility(s) of CaF 2 in an aqueous solution of 0.10 M NaF? CaF 2 (s) <=> Ca 2+ (aq)+ 2F - (aq) K sp =4x10-11 = s (0.10) 2 s = M [Ca 2+ ] = M [F - ]=0.10M. What is the molar solubility (s) of CaF 2 in an aqueous solution of 0.10 M CaCl 2? a. Chemical equilibrium reaction: CaF 2 (s) <=> Ca 2+ (aq)+ 2F - (aq) K sp =4x10-11 = (2s) 2 (0.10) s = (4x10-11 /(4*0.1)) ½ = M [Ca 2+ ] = 0.10M [F - ]= M 4. If you mix 0.5L of 0.1M NaF with 0.5L of 0.1M CaCl 2, there is the possibility of CaF 2 (s) forming. Q sp = [F - ] 2 [Ca 2+ ]=( ) ( ) 2 = >K sp a. With regards to CaF 2, circle the correct choice Q sp < K sp Q sp = K sp Q sp > K sp b. With regards to CaF 2, circle the correct choice Precipitation will occur 8

9 5. The number of moles of a solid that dissolves in 1 liter of water is called the molar solubility. For Ag 2 SO 4, the molar solubility is M. What is the K sp for this compound? a. Chemical reaction: 2Ag + (aq) + SO 4 2- (aq) Ag 2 SO 4 (S) b. Construct an ICE table: [Ag + ]=2s [SO 2-4 ]=s c. K sp= [Ag + ] 2 [SO 4 2- ]= (( ) 2) 2 ( )= A solution is prepared from 0.02 mol of MgCl 2 and mol of NaOH in 1L of water. What is the ph of the solution? The K sp of Mg(OH) 2 is 6 x (Answer:9.) Mg(OH) 2 (s) 2OH (aq) + Mg 2+ (aq) I: 0.004mol (LR) 0.02mol excess Q sp >K sp precipitation forms first C: Revised Initials: mol/1L C +2s +s (s is very small)) Eq 2s= [OH ] [Mg 2+ ] = 0.018M K sp = = [OH ] 2 [Mg 2+ ] =(2s) s= [OH ]= poh=4.74 ph= An 80. ml sample of M lead nitrate, Pb(NO ) 2, is titrated with M sodium fluoride, NaF, until the first appearance of precipitate. The precipitate first appears after 20. ml of sodium fluoride are added. You can assume that the total final volume is 100. ml. What is the value of K sp for PbF 2? (Answer: 1.*10-7 ) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [F ] 2 [Pb 2+ ]= [Pb 2+ ]= 210 M 0.1 [F ] = 210 M The molar solubility of silver chloride is 4.0x10 5. Calculate the K sp of silver chloride.(answer: 16*10-10 ) [Ag + ]=s [Cl - ]=s K sp= [Ag + ][Cl - ]= (( ) 2 =

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Write the chemical reaction(s) that will precede 100%. (Hint: limiting reagent)

Write the chemical reaction(s) that will precede 100%. (Hint: limiting reagent) Chemistry 102 20118 Discussion #8, Chapter 14 and 15 http://quantum.bu.edu/courses/ch102-spring-2018/notes/acidbaseoverview.pdf Student name TA name Section 1. If 1.0 liter solution has 5.6mol HCl, 5.3mol