Really useful information = H + = K w. K b. 1. Calculate the ph of a solution made by pouring 5.0 ml of 0.20 M HCl into 100. ml of water.

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1 Acid Base Equilibrium Putting it all together HA H + + A H + A incomingsa HA +incomingsa Strong Acids HCl HNO3 HBr H2SO4 HI HClO4 HClO3 Really useful information K w H + OH K w M V M V B + H2O OH + HB + OH + +incomingsa HB -incomingsb incomingsa B Strong Bases LiOH CsOH NaOH Ca(OH)2 KOH Sr(OH)2 RbOH Ba(OH)2 1. Calculate the ph of a solution made by pouring 5.0 ml of 0.20 M HCl into 100. ml of water. 2. Calculate the ph of a solution made by pouring 10.0 ml of 0.20 M NaOH into 100. ml of water. 3. Calculate the ph of 100. ml of 0.20 M nitrous acid, HNO2 (Ka ). 4. Calculate the ph of 100. ml of 0.20 M ammonia, NH3 (Kb ). 5. Calculate the ph of a solution made by dissolving 1.38 g of NaNO2 (MM 69.0 g/mol) into enough water to produce 100. ml of solution. 6. Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem Calculate the ph of a solution produced by dissolving 1.38 g of NaNO2 into 100. ml of 0.20 M nitrous acid, HNO2. (Assume that there is no volume change.) 9. Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem Calculate the resulting ph after adding 110. ml of 0.20 M HCl into the solution from problem Calculate the resulting ph after adding 120. ml of 0.20 M NaOH into the solution from problem Calculate the resulting ph after adding 100. ml of 0.20 M HCl into the solution from problem Calculate the resulting ph after adding 100. ml of 0.20 M NaOH into the solution from problem Calculate the resulting ph after adding ml of 0.22 M NaOH into the solution from problem Calculate the resulting ph after adding ml of 0.22 M HCl into the solution from problem Calculate the resulting ph after adding 10.0 ml of 0.20 M HCl into the solution from problem Calculate the resulting ph after adding 20.0 ml of 0.20 M NaOH into the solution from problem 4.

2 Salts & Buffers Putting it all together 1. For (a) (g) which of the following salts, when dissolved in water, at 25ºC, cause a change in ph? If there is a ph change, will the ph be above or below 7? Write a hydrolysis reaction that illustrates the cause of the ph change. Calculate the ph if 2.0 g of the salt is dissolved in 100 ml of water. For (h) (v) will the ph be 7, above 7, or below 7? Justify. a. barium chloride, BaCl2 (208.2 g/mol) b. ammonium perchlorate, NH4ClO4 ( g/mol) c. sodium cyanide, NaCN (49.01 g/mol) Equilibrium Konstants Ka or Kb as appropriate for the listed substance. HCN Ka HF Ka HC2H3O2 Al 3+ Ka Ka NH3 Kb (C2H5)2NH Kb K s for H3PO4 Ka Ka Ka l. equal volumes of 1 M NH3 and 1 M NH4NO3 m. 50 ml of 1 M NH3 and 25 of 1 M HNO3 n. 50 ml of 1 M NaOH and 25 of 1 M HNO3 d. ammonium nitrate, NH4NO3 ( g/mol) o. 100 ml of 1 M NH3 and 50 ml of 1 M NaOH e. potassium acetate, KC2H3O2 ( g/mol) p. equal volumes of 1 M NaBr and 1 M KCl f. diethylamide bromide, (C2H5)2NH2Br ( g/mol) q. equal volumes of 1 M KOH and 1 M HClO4 g. aluminum perchlorate, Al(ClO4)3 ( g/mol) r. equal volumes of 1 M CsOH and 1 M HF h. equal volumes of 1 M KCl and 1 M HCl s. equal volumes of 1 M HC2H3O2 and 2 M NaOH i. equal volumes of 1 M HF and 1 M HCl t. equal volumes of 1 M CsF and 2 M HCl j. equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2 u. equal volumes of 1 M H3PO4 and 1 M NaOH k. equal volumes of 1 M HF and 0.5 M NaOH v. equal volumes of 1 M NaH2PO4 and 1 M Na2HPO4

3 Acid Base Equilibrium Putting it all together 1. Calculate the ph of a solution made by pouring 5.0 ml of 0.20 M HCl into 100. ml of water. M V M V ( 0.2M )( 5ml) ( M )( 105ml) M M ph 2.02 This is simply a SA that has been diluted, need to calculate it s concentration. 2. Calculate the ph of a solution made by pouring 10.0 ml of 0.20 M NaOH into 100. ml of water. M V M V 0.2M ( 10ml) ( M )( 110ml) M 0.018M poh 1.74 ph This is simply a SB that has been diluted, need to calculate it s concentration. 3. Calculate the ph of 100. ml of 0.20 M nitrous acid, HNO2 (Ka ). H + A incomingsa x2 HA +incomingsa [ 0.2] H M ph 2.02 This is a WA, nothing done to it. This is an problem. 4. Calculate the ph of 100. ml of 0.20 M ammonia, NH3 (Kb ). OH HB + +incomingsa x2 B incomingsa [ 0.2] This is a WB, nothing done to it. This is an problem. OH M poh 2.02 ph Calculate the ph of a solution made by dissolving 1.38 g of NaNO2 (MM 69.0 g/mol) into enough water to produce 100. ml of solution. OH HB + +incomingsa B incomingsa [ cjwb] mol Vol x2 0.2 [ ] 1.38g 1mol 69g [ cjwb] 0.1L K w M OH M poh 5.68 ph

4 Acid Base Equilibrium Putting it all together This is a WB with its WA buffer but nothing done to it no incoming SA or SB. You can calculate [OH ] or [H + ] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in 5. mmolcjwb 1.38g 1mol 69g OH HB + +incomingsa B incomingsa H + A incomingsa HA +incomingsa 0.02mol 20mmolcjWB OH H M mmolWA OH M poh ph 3.35 H M ph 3.35 This is a WB with its WA buffer but this time with incoming SB. You can calculate [OH ] or [H + ] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in 5. mmolcjwb 1.38g 1mol 69g OH HB + +incomingsa B incomingsa H + A incomingsa HA +incomingsa 0.02mol 20mmolcjWB OH H M mmolWA 0.2M 10 2mmolSB OH M poh ph 3.43 H M ph 3.43 This is a WB with its WA buffer but this time with incoming SA. You can calculate [OH ] or [H + ] from the base or acid perspective. If calculating from the base perspective, you must calculate Kb from the Ka of HNO2, as done in 5. mmolcjwb 1.38g 1mol 69g OH HB + +incomingsa B incomingsa H + A incomingsa HA +incomingsa 0.02mol 20mmolcjWB OH H M mmolWA 0.2M 20 4mmolSA OH M poh ph 3.17 H M ph 3.17 This is a WB with its WA buffer but this time with incoming SA, in which we have gone 2 mmol beyond the buffer capacity. Anytime we go beyond the buffer capacity, or beyond an equivalence point. It is only about the excess incoming strong (acid or base) and total volume to calculate the molarity of the excess. mmolwb 1.38g 1mol 69g 0.02mol 20mmolWB H + 2mmol M ph ml 0.2M mmolWA 0.2M mmolSA

5 Acid Base Equilibrium Putting it all together NaNO2 is a soluble alkali salt that puts a weak base, NO 2 ion into solution, nothing else has been done to the weak base ion, thus, this is an problem. 6. Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem 3. M V mol ( 0.2M )( 100ml) 20mmolWA ( 0.2M )( 10ml) 2mmol incomingsb H + A incomingsa H + ~ HA +incomingsa ( 20 2) H M ph 2.39 This is a WA with incoming SB, not at halfway, not yet at the equivalence point. 7. Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem 4. OH cjwa +incomingsa WB incomingsa OH ~ OH M poh 4.14 ph OH ~ OH M poh 4.74 ph 9.26 This is a WB with incoming SA, not at halfway, not yet at equivalence point. WB, incoming SA...What if we had added 50 ml? It would be halfway to equivalence point, which is a beautiful thing!

6 Acid Base Equilibrium Putting it all together 8. Calculate the ph of a solution produced by dissolving 1.38 g of NaNO2 into 100. ml of 0.20 M nitrous acid, HNO2. (Assume that there is no volume change.) 9. Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem Calculate the resulting ph after adding 110. ml of 0.20 M HCl into the solution from problem 8.

7 Acid Base Equilibrium Putting it all together 12. Calculate the resulting ph after adding 120. ml of 0.20 M NaOH into the solution from problem 8. Equilibrium Konstants Ka or Kb as appropriate for mmolwb 1.38g 1mol the listed substance. 0.02mol 20mmolWB 0.2M mmolWA 0.2M mmolSB 69g HCN Ka OH 4mmol HF Ka M poh 1.74 ph HC2H3O2 Ka ml 5 This is a WB with its WA buffer but this time with incoming SB, in which we have gone 4 mmol beyond Al 3+ the buffer Ka capacity Anytime we go beyond the buffer capacity, or beyond an equivalence point. It is only about the excess NH3 incoming Strong Kb (acid or 5 base) and total volume to calculate the molarity of the excess. (C2H5)2NH Kb Calculate the resulting ph after adding 100. ml of 0.20 M HCl into the solution from problem 4. K s for H3PO4 Ka Ka M mmolWB 0.2M mmol incomingsa K w [ cjwa] cjwa [ ] 20mmol cjwa 200ml totalvol 0.10M x2 0.1 [ ] H M ph 5.12 WB with incoming SA, and we ve reached the equivalence point! Now it is an problem for the cjwb swimming in the total volume. 14. Calculate the resulting ph after adding 100. ml of 0.20 M NaOH into the solution from problem M mmolWA 0.2M mmol incomingsb K w [ cjwb] cjwb [ ] 20mmol cjwb 200ml totalvol 0.10M x2 0.1 [ ] OH M poh 5.83 ph 8.17

8 Acid Base Equilibrium Putting it all together WA with incoming SB, we ve reached the equivalence point! Now it is an problem for the WA swimming in the total volume. 15. Calculate the resulting ph after adding ml of 0.22 M NaOH into the solution from problem M mmolWA 0.22M mmolSB OH 2mmol 0.01M poh 2.00 ph ml This is a WA with incoming SB, in which we have gone beyond 2 mmol beyond the equivalence point. Anytime we go beyond the equivalence point, it is only about the excess incoming SB and total volume to calculate the molarity of the excess. 16. Calculate the resulting ph after adding ml of 0.22 M HCl into the solution from problem M mmolWB 0.22M mmolSA H + 2mmol 0.01M ph ml This is a WB with incoming SA, in which we have gone beyond 2 mmol beyond the equivalence point. Anytime we go beyond the equivalence point, it is only about the excess incoming SA and total volume to calculate the molarity of the excess. 17. Calculate the resulting ph after adding 10.0 ml of 0.20 M HCl into the solution from problem 3.

9 Acid Base Equilibrium Putting it all together The trick to the following questions is to be on the lookout for the SA pathetics, negative ions of SA (Cl, Br, I, NO3, SO4, ClO4, ClO3 ), and the SB pathetics, Group 1 & 2 positive ions (Li +, Na +, K +, Rb +, Cs +, Ba 2+, Sr 2+, Ca 2+ ). If you know who the pathetics are, then all other positive ions are conjugate H + 2mmolSA weak acids, and (nearly) all other negative + ions 0.018M ph 1.74 H 110totalVol 2.95mmolSA are conjugate weak bases M ph totalVol K s for H3PO4 Ka Ka Ka WA with SA added to it. This is a curveball, AP is not likely to ask this question, but they have once before...so just in case. The WA is unimportant compared to the amount of SA coming in, while it is tempting to add the mmol of WA present from problem KCl is made 3 (0.95 of two mmol), pathetic the ions, incoming and HCl acid is would a SA. Thus actually this suppress NaBr the amount is made of of H two + contributed pathetic ions, by the Na WA; + and think Br, and KCl is LeChatelier s is NOT a buffer principle. and is has (Likely a ph the below ph 7. actually somewhere between also 1.74 made and 1.57) out of two pathetics, K + and Cl. Thus this is NOT a buffer just lots of non-hydrolyzing ions, thus the ph is 7. HF is a WA, and HCl is a SA, thus this is nothing more than lots of acid; NOT a buffer, just ph below Calculate the resulting ph after adding 20.0 ml of 0.20 M NaOH KOH into the is made solution of a from SB, and problem 4. HClO4 is a SA. Since the SA and SB are present in equal quantities, neutralization occurs (H2O is formed) and all that remains are non-hydrolyzing ions (K + and ClO4 ) thus the ph is 7. HC2H3O2 is a WA, and for NaC2H3O2, the Na + is a pathetic and the C2H3O2 is the conjugate WB of the WA, HC2H3O2. Since both the WA and WB are present in equal quantities, this is a buffer OH 4mmolSB with ph below 7. The ph will be ~4ish M poh 1.74 ph totalVol OH 4.19mmolSA 0.035M poh 1.46 ph CsOH is made of a SB, and HF is a WA. Since the SB and 120totalVol WA are present in equal quantities, neutralization occurs WB with SB added to it. This is a curveball, AP is not likely to ask this (H2O question, is formed) but they and what have remains once before...so in the beaker just in is case. a pathetic, The HF is WB a WA, is unimportant and NaOH compared is a SB. Look to the at amount the molarities of SB to coming se in, while Cs +, it and is tempting a WB, F to, thus add the ph mmol is above of WB 7. present from problem that the SB 4 will (0.19 neutralize mmol), the half incoming the WA, thus base half would of the suppress WA the amount of OH contributed by the WB; think LeChatelier s principle. will be left, (Likely and half ph of actually the WA, somewhere HF, will turn between into WB, F and ) This is an acidic buffer with ph below 7. The ph will be ~3ish NH3 is a WB, and NH4NO3 is a pathetic, NO3, and the conjugate WA, NH4 +, of the WB, NH3. Since both the WB and WA are present in equal quantities, this is a basic buffer with ph above 7. The ph will be ~9ish NH3 is a WB, and HNO3 is a SA. Since the SA will neutralize half the WB, half of the WB will be left, and half of the WB will turn into WA. This is a basic buffer with ph above 7. The ph will be ~9ish NaOH is a SB and HNO3 is a SA. We have twice as much base as acid, but since there is no WA and no WB, this is NOT a buffer. It is simply more SB present than SA, thus the ph will be above 7. NH3 is a WB, and NaOH is a SB, thus this is nothing more than lots of base; NOT a buffer, just ph above 7. HC2H3O2 is made of a WA, and NaOH is a SB. Since the SB is present in twice the quantity needed to neutralize the WA, there is lots of excess SB left over, thus the solution will NOT be a buffer with a ph above 7. CsF is made of a pathetic and a WB, and HCl is a SA. Since the WB and SA are present in equal quantities, neutralization occurs (H2O is formed) and what remains in the beaker is are two pathetics, Cs + and Cl, and a WA, HF. Thus the ph is below 7. H3PO4 is made of a WA, and NaOH is a SB. Since enough SB is present to neutralize one of the ionizable H + ions, H2PO4, will remain in the beaker. This is a WA, with very little of it s conjugate WB present and is thus NOT a buffer, but will be acidic with a ph below 7. NaH2PO4 is made of a pathetic, Na +, and a WA, H2PO4. Na2HPO4 is a is a pathetic, Na +, and a WB, HPO4 2 (the WB of H2PO4 ). This is a buffer, and would likely have a ph ~7ish, the pka of the WA, H2PO4.

10 Salts & Buffers Putting it all together 1. Which of the following salts, when dissolved in water, at 25ºC, cause a change in ph? If there is a ph change, will the ph be above or below 7? Write a hydrolysis reaction that illustrates the cause of the ph change. Calculate the ph if 2.0 g of the salt is dissolved in 100 ml of water. a. barium chloride, BaCl2 (208.2 g/mol) ph will remain at 7. BaCl2 Ba Cl (note the single arrow since BaCl2 is a soluble ionic compound). The Ba 2+ ions are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. The Cl ions are also pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. b. ammonium perchlorate, NH4ClO4 ( g/mol) 2g 1mol g 0.017mol 0.017mol 0.17M 0.1L K w [ ] [ ] H + NH 3 + NH 4 x H ph 5.01 ph will change to below 7, acidic. NH4ClO4 NH4 + + ClO4 (note the single arrow since NH4NO3 is a soluble ionic compound) The ClO4 ions are are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. However, the ammonium ions are a WA, thus the NH4 + ions partially dissociate in water and produce H + ions. NH4 + H + + NH3 c. sodium cyanide, NaCN (49.01 g/mol) 2g 1mol 49.01g mol mol 0.408M K w 0.1L [ ] x OH poh 5.47 ph [ ] OH HCN CN

11 Salts & Buffers Putting it all together ph will change to above 7, basic. NaCN Na + + CN (note the single arrow since NaCN is a soluble ionic compound) The Na + ions are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. But the CN ions are a conjugate WB which hydrolyzes in water and produces extra OH ions. CN + H2O OH + HCN d. ammonium nitrate, NH4NO3 ( g/mol) 2g 1mol g 0.025mol 0.025mol 0.250M K w 0.1L [ ] x H ph [ ] H + NH 3 + NH 4 ph will change to below 7, acidic. NH4NO3 NH4 + + NO3 (note the single arrow since NH4NO3 is a soluble ionic compound) The NO3 ions are are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. But the NH4 + ions are a conjugate WA and partially dissociate in water and produce H + ions. NH4 + H + + NH3

12 Salts & Buffers Putting it all together e. potassium acetate, KC2H3O2 ( g/mol) 2g 1mol g mol mol 0.204M K w 0.1L [ ] ph will change to above 7, basic. KC2H3O2 K + + C2H3O2 (note the single arrow since KC2H3O2 is a soluble ionic compound) The K + ions are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. But the C2H3O2 ions are a conjugate WB which hydrolyzes in water and produces extra OH ions. C2H3O2 + H2O OH + HC2H3O2 x OH poh 4.97 ph [ ] OH HC 2 H 3 O 2 C 2 H 3 O 2 f. diethylamide bromide, (C2H5)2NH2Br ( g/mol) 2g 1mol g 0.013mol 0.013mol 0.130M K w 0.1L [ ] ph will change to below 7, acidic. (C2H5)2NH2Br (C2H5)2NH2 + + Br (note the single arrow since (C2H5)2NH2Br is a soluble ionic compound, you should be able to recognize this because of the Br at the end of the molecule and the acid/base context of the question) The Br ions are are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. But the (C2H5)2NH2 + ions partially dissociate in water and produce H + ions. (C2H5)2NH2 + H + + (C2H5)2NH x H ph [ ] H + (C 2 H 5 ) 2 NH + (C 2 H 5 ) 2 NH 2 g. aluminum perchlorate, Al(ClO4)3 ( g/mol)

13 Salts & Buffers Putting it all together 2g 1mol g mol mol M 0.1L H + Al(OH ) Al [ ] x H ph 3.02 ph will change to below 7, acidic. Al(ClO4)3 Al 3+ + ClO4 (note the single arrow since ClO4 is a soluble ionic compound, you should be able to recognize this because of the ClO4 at the end and the acid/base context) The Al 3+ ions interact with water and produce H + ions. But the ClO4 ions are are pathetic meaning they do not hydrolyze in water and thus do not produce any extra H + or OH ions. Al(H2O)6 3+ Al(H2O)5(OH) 2+ + H + (Note: You would probably never have to write this equation.)

14 Salts & Buffers Putting it all together State whether the ph of the following solutions will be greater than, less than, or equal to 7, and determine which of the following combinations would produce an effective buffer? h. equal volumes of 1 M KCl and 1 M HCl p. equal volumes of 1 M NaBr and 1 M KCl i. equal volumes of 1 M HF and 1 M HCl q. equal volumes of 1 M KOH and 1 M HClO4 j. equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2 r. equal volumes of 1 M CsOH and 1 M HF k. equal volumes of 1 M HF and 0.5 M NaOH s. equal volumes of 1 M HC2H3O2 and 2 M NaOH l. equal volumes of 1 M NH3 and 1 M NH4NO3 t. equal volumes of 1 M CsF and 1 M HCl m. 50 ml of 1 M NH3 and 25 of 1 M HNO3 u. equal volumes of 1 M H3PO4 and 1 M NaOH n. 50 ml of 1 M NaOH and 25 of 1 M HNO3 v. equal volumes of 1 M NaH2PO4 and 1 M Na2HPO4 o. 100 ml of 1 M NH3 and 50 ml of 1 M NaOH