Solutions to: Acid-Base and Solubility Homework Problem Set S.E. Van Bramer 1/3/17

Transcription

1 Solutions to: Acid-Base and Solubility Homework Problem Set S.E. Van Bramer /3/7.Calculate the H 3 O + concentration, OH - concentration, ph, and poh of the following solutions. First solve assuming that K w is insignificant. Repeat the calculations and include K w (you will need to use the quadratic equation to solve these). When is K w significant? : liter K w : a..0 x 0-9 HNO 3 C acid :.00 9 Assuming K w is insignificant: C H3O : C acid C H3O 0 9 ph : log C H3O ph 9 Does this make sense? poh : 4 ph poh 5 Assuming K w is significant (Included the initial [H 3 O + ] in the calculations.) C H3O : K w + C acid C H3O ph : log C H3O ph poh : 4 ph poh 7.004

2 -b..0 x 0-6 HNO 3 C acid :.00 6 Assuming K w is insignificant: C H3O : C acid C H3O 0 6 ph : log C H3O ph 6 poh : 4 ph poh 8 Assuming K w is significant (Included the initial [H 3 O + ] in the calculations.) C H3O : K w + C acid C H3O. 0 6 ph : log C H3O ph poh : 4 ph poh c.0.0 HNO 3 C acid : 0 Assuming K w is insignificant: C H3O : C acid C H3O 0 ph : log C H3O ph poh : 4 ph poh 5 Notice that ph can be negative. A t concentrations this high, however, the assumption that all HNO 3 dissociates is probably not accurate. Assuming K w is significant (Included the initial [H 3 O + ] in the calculations.) C H3O : K w + C acid C H3O 0 ph : log C H3O ph poh : 4 ph poh 5

3 . Calculate the H 3 O + concentration, OH - concentration, ph, poh, nitrate ion concentration and sodium ion concentration for the following experimental steps in a titration. Graph your results. -a.6 HNO 3. This is just a strong acid. C acid : 6 C H3O : C acid C H3O 6 ph : log C H3O ph poh : 4 ph poh b..00 ml of the nitric acid is diluted with deionized water to a volume of 00.0 ml. The concentration of the acid is : C stock : 6 V stock :.00mL V acid : 00mL C stock V stock C acid : V acid C acid 0.06 The concentrations: C H3O : C acid C H3O 0.06 ph : log C H3O ph. poh : 4 ph poh.778

4 -c g of NaOH is dissolved in deionized water to a volume of.000 liter. The concentration of sodium hydroxide is: mass base : 5.468gm W base : ( )gm mass base base : W base volume base : 000mL C base : base volume base C base 0.3 The concentrations: C OH : C base C OH 0.3 poh : log C base poh 0.88 ph : 4 poh ph 3.8 -d..00 ml of the sodium hydroxide solution is added to the nitric acid solution. V acid : 00mL V base :.00mL : V acid + V base 0mL Calculate the ph by assuming that all the added OH-, reacts with H3O already present. H3O : C acid V acid C base V base H3O H3O C H3O : C H3O C OH K w : C OH C H3O ph : log C H3O ph.36 poh : 4 ph poh.764 C acid V acid C NO3 : C NO C base V base C Na : C Na

5 -e.0.00 ml of the sodium hydroxide solution is added to the nitric acid solution. V acid : 00mL V base : 0mL : V acid + V base 0mL Calculate the ph by assuming that all the added OH-, reacts with H3O already present. H3O : C acid V acid C base V base H3O H3O C H3O : C H3O 0.08 C OH K w : C OH C H3O ph : log C H3O ph.55 poh : 4 ph poh.449 C acid V acid C NO3 : C NO C base V base C Na : C Na 0.0

6 -f.enough sodium hydroxide solution is added to reach the equivilence point. At the equivalence point enough base is added so that the s of hydroxide and the s of nitric acid are equivilent. acid : C acid V acid acid base : acid base base V base : V base mL C base : V acid + V base mL Calculate the ph by assuming that all the added OH-, reacts with H3O already present. H3O : C acid V acid C base V base H3O 0 Since the acid and base do not contribute to the ph, it is the same as pure water: For pure water, the K w equlibrium determines the ph K w C H3O C OH C H3O X C OH X X : K w C H3O : X C H3O 0 7 ph : log C H3O ph 7 poh : 4 ph poh 7 C acid V acid C NO3 : C NO C base V base C Na : C Na 0.04

7 -g..00 ml of the sodium hydroxide solution is added to the nitric acid solution. V acid : 00mL V base : 50mL : V acid + V base 50mL Calculate the ph by assuming that all the added OH-, reacts with H3O already present. H3O : C acid V acid C base V base H3O This result does not make any sense because nitric acid is now the limiting reagent. So OH - is in excess and determines the ph OH : C base V base C acid V acid OH OH C OH : C OH poh : log C OH poh.49 ph : 4.0 poh ph.57 C acid V acid C NO3 : C NO C base V base C Na : C Na 0.044

8 3.What is the ph and poh of a buffer prepared by adding.435 g of sodium acetate to 00 ml of 0.4 acetic acid. Calculate using the Henderson-Hasselbalch equation (as shown in your textbook) and using the quadratic equation (as shown in class). K a :.80 5 The concentration of the acid in the buffer is: C acid : 0.4 The concentration of the salt is (Fill in appropriate values and delete unnecessary sections.): mass salt :.435gm W salt : [[(.990) + (.00 ) + ( )] + ( )]gm mass salt salt : salt 0.05 W salt V salt : 00mL salt C salt : C salt 0.5 V salt The equlibrium solution for HA + HO <-> H3O+ + A- Initial Concentraion C acid C salt Equlibrium Concentration C acid X X C salt + X Equlibrium expression: ( C salt + X)X K a C acid X Assuming X is smaller than [acid] and [salt], this reduces to: Solving for X: ( ) X ( ) C salt ( ) K a X : C acid K a C acid C salt X

9 Equlibrium Concentrations: C HA : C acid X C HA 0.4 C H3O : X C H3O C A : C salt + X C A 0.5 ph : log C H3O ph 4.83 poh : 4 ph poh 9.68 Solving for X without assumptions (You will need to use the quadratic equation): ( C salt + X)X K a C acid X X : K a K a C salt + C salt K a + K a C salt + C salt + 4K a K a + K a C salt + C salt + 4K a C acid C acid X Equlibrium Concentrations (Calculated for both roots, select the reasonable answer [top or bottom in ()]: 0.4 C HA : C acid X C HA C H3O : X C H3O C A : X C A 0.5 i : 0,.. ph : log C i H3Oi ph poh : 4 ph poh i i

10 4.Determine the ph and poh of a buffer prepared by adding g of sodium hydroxide to 00 ml of 0.4 acetic acid. Calculate using the Henderson-Hasselbalch equation (as shown in your textbook), the method of successive approximations, and using the quadratic equation (as shown in class). Given a buffer prepared from a weak acid where (Input appropriate value here): K a :.80 5 The concentration and volume of the acid is: C acid : 0.4 V acid : 00mL Since sodium hydroxide is a strong base it will react completely with the strongest available acid (the acetic acid). This reaction will convert acetic acid into acetate. The s of acid present initially acid : C acid V acid acid 0.0 The s of sodium hydroxide added mass base : 0.000gm W base : ( )gm mass base base : W base base Since this reaction goes to completion: acetate : base acetate mol acid : acid base acid 0.0mol acid C acid : V acid acetate C acetate : V acid

11 The equlibrium solution for HA + HO <-> H3O+ + A- Initial Concentraion C acid C acetate Equlibrium Concentration C acid X X C acetate + X Equlibrium expression: ( C acetate + X)X K a C acid X Assuming X is smaller than [acid] and [salt], this reduces to: Solving for X: ( ) X ( ) C acetate ( ) K a X : C acid K a C acid C acetate X Equlibrium Concentrations: C HA : C acid X C HA 0. C H3O : X C H3O C A : C acetate + X C A ph : log C H3O ph poh : 4 ph poh 0.94

12 Solving for X without assumptions: ( C salt + X)X K a C acid X X : K a K a C acetate + C acetate K a + K a C acetate + C acetate + 4K a K a + K a C acetate + C acetate + 4K a C acid C acid X Equlibrium Concentrations (Calculated for both roots, select the reasonable answer [top or bottom in ()]: 0. C HA : C acid X C HA C H3O : X C H3O C A : X C A i : 0,.. ph : log C i H3Oi ph poh : 4 ph poh i i

13 5.Calculate the H 3 O + concentration, OH - concentration, ph, poh, acetic acid concentration, and the acetate ion concentration for the following experimental steps in a titration. Calculate using the Henderson-Hasselbalch equation (as shown in your textbook) and using the quadratic equation (as shown in class). Graph your results. a..00 ml of glacial ac etic acid (pure ac etic acid, density.049 g/ml) is diluted with deionized water to a volume of 00.0 ml. Starting with the following solution of acetic acid and determine the initial acid concentration: K a : V acid :.00mL density acid :.049gm ml mass acid : V acid density acid W acid : [(.00 ) + ( ) + ( )]gm mass acid acid : acid 0.07 W acid V acid : 00.0mL acid C acid : C acid 0.75 V acid The equlibrium solution for HA + HO <-> H3O+ + A- X K a C acid Solving for X: X : K a C acid X Equlibrium Concentrations: C HA : C acid X C HA 0.73 C H3O : X C H3O C A : X C A ph : log C H3O ph.75 poh : 4 ph poh.49

14 Solving for X without assumptions: X K a C acid X X : K a K a + K a + 4K a K a + 4K a C acid C acid X Equlibrium Concentrations (Calculated for both roots, select the reasonable answer [top or bottom in ()]: 0.73 C HA : C acid X C HA C H3O : X C H3O C A : X C A i : 0,.. ph : log C i H3Oi ph poh : 4 ph poh i i

15 5-b..468 g of NaOH is dissolved in deionized water to a volume of 500 ml. The concentration of the base is: mass base :.468gm W base : ( )gm mass base base : W base volume base : 500mL C base : base volume base C base 0.3 The concentrations: C OH : C base C OH 0. poh : log C base poh ph : 4 poh ph 3.05

16 5-c. 0 ml of the sodium hydroxide solution is added to the acetic acid solution. This shifts the equlibrium from the above system. To solve for the new equlibrium conditions, two steps are required. First since OH - is a strong base, and acetic acid is the strongest acid available, CH 3 COOH + OH - -> CH 3 COO - This reaction will go to completion so that for: C NaOH : C base V NaOH : ml + H O NaOH : C NaOH V NaOH NaOH : V acid + V NaOH 0.0liter The new initial, NON-EQULIBRIU, conditions for acetic acid and acetate are: Acetic acid Acetate Ions CH3COOH : acid NaOH CH3COO : NaOH C CH3COOH ( ) CH3COOH CH3COO : C CH3COO : ( ) C CH3COOH C CH3COO. 0 3 Based upon these initial concentrations, solve for the equlibrium values, assuming X reacts: ( C CH3COO + X)X K a C CH3COOH X Assuming X is small, we may simplify using the Henderson Hasselbach approximation C CH3COO X K a Rearanges to: X : CH CH3COOH Which solves as: X K a C CH3COOH C CH3COO This gives the following equlibrium concentraitons: C HA : C CH3COOH X C HA 0.69 C A : C CH3COO + X C A C H3O : X C H3O ph log C H3O : ph.556 poh : 4 ph poh.444

17 If [H3O+] is smaller than the buffer concentrations, the answer is acceptable. In this example the approximation is not very good. Without making the Henderson Hasselbach approximation, the above problem solves as: ( C CH3COO + X)X K a C CH3COOH X Has solutions (using the quadratic equation) X : K a K a C CH3COO C CH3COO + K a + K a C CH3COO + C CH3COO + 4K a K a + K a C CH3COO + C CH3COO + 4K a C CH3COOH C CH3COOH X Selecting the appropriate root: (NOTE: this answer is significantly different from above.) This gives the following equlibrium concentraitons: 0.74 C HA : C CH3COOH X C HA C A : C CH3COO + X C A C H3O : X C H3O C H3Oi ph : log ph i poh : 4 ph poh i i.08

18 5-e.0.00 ml of the sodium hydroxide solution is added to the acetic acid solution. V NaOH : 0mL NaOH : C NaOH V NaOH NaOH : V acid + V NaOH 0.liter The new initial, NON-EQULIBRIU, conditions for acetic acid and acetate are: Acetic acid Acetate Ions CH3COOH : acid NaOH CH3COO : NaOH C CH3COOH ( ) CH3COOH CH3COO : C CH3COO : C CH3COOH C CH3COO 0.0 ( ) Based upon these initial concentrations, solve for the equlibrium values, assuming X reacts: ( C CH3COO + X)X K a C CH3COOH X Assuming X is small, we may simplify using the Henderson Hasselbach approximation C CH3COO X K a Rearanges to: X : CH CH3COOH Which solves as: X K a C CH3COOH C CH3COO This gives the following equlibrium concentraitons: C HA : C CH3COOH X C HA 0.48 C A : C CH3COO + X C A 0.0 C H3O : X C H3O ph log C H3O : ph 3.58 poh : 4 ph poh 0.48

19 If [H3O+] is smaller than the buffer concentrations, the answer is acceptable. In this example the approximation is not very good. Without making the Henderson Hasselbach approximation, the above problem solves as: ( C CH3COO + X)X K a C CH3COOH X Has solutions (using the quadratic equation) X : K a K a C CH3COO C CH3COO + K a + K a C CH3COO + C CH3COO + 4K a K a + K a C CH3COO + C CH3COO + 4K a C CH3COOH C CH3COOH X Selecting the appropriate root: (NOTE: this answer is significantly different from above.) This gives the following equlibrium concentraitons: 0.59 C HA : C CH3COOH X C HA C A : C CH3COO + X C A C H3O : X C H3O C H3Oi ph : log ph i poh : 4 ph poh i i 0.407

20 5-e.Enough sodium hydroxide solution is added to reach the equivilence point. At the equivilance point the same number of s of sodium hydroxide have been added as there are s of acetic acid to begin with. acid 0.07mol NaOH : acid NaOH V NaOH : V NaOH mL C base : V acid + V NaOH liter The new initial, NON-EQULIBRIU, conditions for acetic acid and acetate are: Acetic acid Acetate Ions CH3COOH : acid NaOH CH3COO : NaOH C CH3COOH ( ) CH3COOH CH3COO : C CH3COO : ( ) C CH3COOH 0 C CH3COO K w K b : K b K a The equlibrium solution for A- + HO <-> OH- + HA Initial Concentraion C CH3COO Equlibrium Concentration C CH3COO X X X Equlibrium expression: X K b C CH3COO X Assuming X is smaller than [acid], this reduces to: X K b C CH3COO

21 Solving for X: X : K b C CH3COO X Equlibrium Concentrations: C HA : X C HA C OH : X C OH C A : C CH3COO X C A poh : log C OH poh 5. ph : 4 poh ph 8.79 Solving for X without assumptions: X K b C CH3COO X K b + K b + 4K b C CH3COO X : X K b K b + 4K b C CH3COO Equlibrium Concentrations (Calculated for both roots, select the reasonable answer [top or bottom in ( ) ]: C HA : X C HA C OH : X C H3O C A : C CH3COO X C A poh : log C i OHi poh ph : 4 poh ph i i

22 5-f.00.0 ml of the sodium hydroxide solution is added to the acetic acid solution. This is an excess of NaOH. Since OH - is the strongest base present it will determine the ph V NaOH : 00mL NaOH : C NaOH V NaOH NaOH 0.0 : V acid + V NaOH 0.3liter First OH - will convert all the CH 3 COOH to CH3COO - these initial conditions are: CH3COOH : 0 CH3COO : acid Calculate the [OH-], from the excess OH. OH : NaOH acid OH OH C OH : C OH 0.07 Then use this to calculate poh and ph: poh log C OH : poh.779 ph : 4 poh ph.

23 6. Assuming the molar solubility of a salt is x, write the balanced chemical equation for dissolving the solid salt, the K sp expression and the K sp expression solved for x. 6-a. AB AB(s) <-> A + B K sp ( X) ( X) X K sp 6-b. AB AB (s) <-> A + B K sp ( X) ( X ) K sp 4X 3 X K sp 6-c. A B A B(s) <-> A + B K sp ( X ) ( X) K sp 4X 3 X 4 K sp d. AB 3 AB 3 (s) <-> A + 3 B K sp ( X) ( 3X ) 3 K sp 7X 4 X K sp e. A B 3 A B 3 (s) <-> A + 3 B K sp ( X ) ( 3X ) 3 K sp 08X 5 X K sp 08 5

24 7. 3. * 0-3 g of maganese (II) hydroxide will dissolve in liter of water at 5 o C. What is the ph of a saturated manganese (II) hydroxide solution? What is K sp for manganese(ii) hydroxide? 7-a. The ph of the solution: First calculate the s of n(oh) dissolved mass : gm gm W : [ ( ) ] W 88.95gm mass noh : noh W Next calculate the concentration of OH - volume : liter : liter noh C OH : C OH volume Then the concentration of H 3 O + K w :.00 4 C H3O K w : C H3O C OH Then the ph ph : logc H3O ph b. Calculate K sp for n(oh) C OH C n : C n And the appropriate expression for K sp K sp ( ) ( C n ) C OH : K sp

25 8. How many grams of silver chromate can dissolve in 50 ml of 0.5 sodium chromate? The equilibrium reaction is: Ag CrO 4 <-> Ag + CrO 4 The initial conditions 0.5 The equilibrium conditions The equilibrium expression: K sp ( X ) ( X) X X For Ag CrO 4 K sp :.0 If we assume that X is much smaller than 0.5, the equilibrium expression simplifies to K sp ( X ) ( 0.5) simplifies to K sp X X : K sp X For 50 ml of solution: AgCrO4 : ( X ) ( 0.50liter ) AgCrO Find the mass of silver chromate. W AgCrO4 : [( ) ( )]gm W AgCrO gm ass AgCrO4 : W AgCrO4 AgCrO4 ass AgCrO gm

26 grams of lead nitrate is dissolved in 50.0 ml of water. When 0 ml of.0 NaCl is added to this solution: What is the mass of the preciptate? What is the equlibrium concentration of all the ions in the solution? Find the s of lead nitrate dissolved: ass PbNO3 : 0.500gm W PbNO3 : [ ( ) + ( )]gm ass PbNO3 ole PbNO3 : W PbNO3 ole PbNO Find the s of sodium chloride added: C NaCl :.0 V NaCl : 0mL V NaCl 0.0liter ole NaCl : C NaCl V NaCl ole NaCl 0.0 The precipitation reaction is: Pb + (aq) + Cl - (aq) <--> PbCl (s) K sp C Pb C Cl V : ( )mL V 0.6 liter ole PbNO3 C Pb : C Pb V ole NaCl C Cl : C Cl V Q : C Pb C Cl Q K sp : Since Q is less than K sp, no precipitate is produced in this reaction.

27 Find the s of lead nitrate dissolved: ass PbNO3 : 0.500gm W PbNO3 : [ ( ) + ( )]gm ass PbNO3 ole PbNO3 : W PbNO3 ole PbNO Find the s of sodium chloride added: C NaCl :.0 V NaCl : 0mL V NaCl 0.0liter ole NaCl : C NaCl V NaCl ole NaCl 0.04 The precipitation reaction is: Pb + (aq) + Cl - (aq) <--> PbCl (s) K sp C Pb C Cl V : ( )mL V 0.7 liter ole PbNO3 C Pb : C Pb V ole NaCl C Cl : C Cl 0.48 V Q : C Pb C Cl Q K sp : Q is greater than K sp so a precipitate is produced

28 Pb + (aq) + Cl - (aq) <--> PbCl (s) Initial Conditions: C Pb C Cl 0.48 Change: X X + X The equlibrium expression is: ( ) ( C Cl X ) K sp C Pb X Substitute in values: K sp C Pb Solve for X C Cl ( X) ( X) ( X) ( X + X ) X X X 3 X i i NOTE: This was a cubic expression (X3), so there are three roots. Only the first one is meaningful, the second two are "complex numbers". The i is the "imaginary" part. It is unlikely that you will be able to solve a cubic expression (unless you have access to a symbolic mathematics program (like mathcad, which I am using here), or have completed several semesters of calculs (and really understood it). So we need a different way to solve this problem. We need to make some assumptions.

29 Back to the equlibrium expression:.70 5 ( X) ( X) The problem is greatly simplifed if we "assume" that X is much smaller than ( X) ( 0.485) X X We can check the assumption by evaluating the orignal expression with this value of X ( ) ( ) This is fairly close, so the approximation was pretty good. We can now improve the calculated value of X by using this value for a successive approximation and substitute it into part of the original expression (instead of the assuming that X is much smaller than to simplify the calculation, we use this value). So Back to the original expression:.70 5 ( X) ( X) ake the substitution:.70 5 And solve for the remaining X: ( ) ( X) ( X) ( ) ( ) ( X) X X This value of X should be a bit better than the first estimate, so we go back and substitute it into the original expression and see how it works: ( ) ( ) Which is a very good agreement. It also compares well with the value calculated by solving the cubic expression.

30 So now we can determine the equlibrium concentrations of all the species and the mass of the precipitate. Remember that: Pb + (aq) + Cl - (aq) <--> PbCl (s) Initial Conditions: C Pb C Cl 0.48 Change: X X From above we found the solution for X X : So the equlibrium concentrations are: C Pb_equlibrium : C Pb X C Pb_equlibrium C Cl_equlibrium : C Cl X C Cl_equlibrium 0.39 The amout of precipitate is: ole PbCl : XV ole PbCl W PbCl : ( )gm ass PbCl : ole PbCl W PbCl ass PbCl 0.358gm