104. A chemist titrated 35.00 ml of a 0.150 mol/l solutio of hypobromous acid, HBrO(aq), K a = 2.8 10 9. Calculate the resultig ph after the additio of 15.00 ml of 0.1000 mol/l sodium hydroxide solutio, (aq). What Is Required? You eed to fid the ph at a certai poit i the titratio usig hypobromous acid ad potassium hydroxide solutios. What Is Give? You kow the volume of the hypobromous acid: HBrO 35.00 ml You kow the cocetratio of the hypobromous acid: chbro 0.1500 mol/l You kow the volume of the sodium hydroxide: 15.00 ml You kow the cocetratio of the sodium hydroxide: c 0.1000 mol/l You are give the acid dissociatio costat for hypobromous acid: K a = 2.8 10 9. Pla Your Strategy Write the balaced equatio for the reactio betwee HBrO (aq) ad (aq). Use the formula c to calculate the amout i moles,, of HBrO(aq) preset before ay base was added. Use the formula c to calculate the amout i moles,, of (aq). Calculate the amout i moles,, of BrO (aq) preset after the base was added. Calculate the amout i moles of ureacted HBrO(aq). Act o Your Strategy Because is a strog base, the reactio essetially goes to completio. HBrO(aq) (aq) + (aq) NaBrO(aq) + H 2 O(l) HBrO c mol 0.1500 35.00 ml 1 L L 1000 ml 5.250 10 mol c mol 0.1000 15.00 ml 1 L L 1000 ml 1.500 10 mol Because the reactio goes to completio ad because the ratio of BrO to is 1:1, for every mole of added, oe mole of HBrO will be coverted ito BrO. Therefore: BrO 3 1.500 10 mol excess HBrO HBrO i BrO formed 3 3 5.250 10 mol 1.500 10 mol 3 3.750 10 mol 264 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
Determie the fial volume of the solutio. Use c to determie the cocetratios of HBrO(aq) ad BrO (aq) i mol/l. Write equatios for ay equilibrium reactios occurrig ad determie whether the solutio has bufferig capacity. Write the equilibrium expressio, substitute i the calculated cocetratios ad solve for the cocetratio of the hydroium io. HBrO 35.00 ml + 15.00 ml 1 L = 50.00 ml 1000 ml 0.05000 L chbro 3 3.75010 mol 0.0500 L 2 7.50010 mol/l c BrO 3 1.50010 mol 0.0500 L 2 3.00010 mol/l HBrO(aq) + H 2 O(l) BrO (aq) + H 3 O + (aq) BrO (aq) + H 2 O(l) HBrO (aq) + OH (aq) The solutio has bufferig capacity. The reactios of HBrO ad BrO reactig with water balace each other ad very little chage actually occurs. Therefore, you ca use the amouts of HBrO ad BrO preset after the reactio with ad the equilibrium costat to fid the cocetratio of hydroium io, H 3 O + preset. [BrO ][H3O ] Ka [HBrO] 9 x 2.8 10 2 7.50010 x 3.00010 9 7.000 10 2 (3.000 10 )( ) 9 2 (2.8 10 )(7.500 10 ) 2 [H 3 O + ] = x = 7.000 10 9 mol/l Uit 4 Part B MHR 265
Calculate ph usig: ph = log [H 3 O + ]. ph = log [H 3 O + ] = log (7.000 10 9 ) = 8.1549 = 8.15 Check Your Solutio The umber of sigificat digits to the right of the decimal is the same as the umber of sigificat digits i the give K a. 266 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
105. A solutio of 50.00 ml of a 0.120 mol/l itrous acid, HNO 2 (aq), is titrated with 0.1000 mol/l of potassium hydroxide, KOH(aq). Determie the resultig ph of the solutio after 11.25 ml of the base has bee added. For the acid, K a = 5.6 10 4. What Is Required? You eed to fid the ph at a certai poit i the titratio usig itrous acid ad potassium hydroxide solutios. What Is Give? You kow the volume of the itrous acid: 50.00 ml HNO2 You kow the cocetratio of the itrous acid: c 0.120 mol/l HNO2 You kow the volume of the potassium hydroxide: KOH 11.50 ml You kow the cocetratio of the potassium hydroxide: ckoh 0.1000 mol/l You kow the acid dissociatio costat: K a = 5.6 10 4 Pla Your Strategy Write the balaced equatio for the reactio betwee HNO 2 (aq) ad KOH(aq). Use the formula c to calculate the amout i moles,, of HNO 2 (aq) preset before ay base was added. Use the formula c to calculate the amout i moles,, of KOH(aq). Calculate the amout i moles,, of NO 2 (aq) preset after the base was added. Calculate the amout i moles of ureacted HNO 2 (aq). Act o Your Strategy Because KOH is a strog base, the reactio essetially goes to completio. HNO 2 (aq) + KOH(aq) KNO 2 (aq) + H 2 O(l) HNO 2 c mol 0.120 50.00 ml 1 L L 1000 ml 6.00 10 mol KOH c mol 0.100 11.25 ml 1 L L 1000 ml 1.125 10 mol Because the reactio goes to completio ad because the ratio of NO 2 to KOH is 1:1, for every mole of KOH added, oe mole of HNO 2 will be coverted ito NO 2. Therefore: NO2 KOH 3 1.125 10 mol excess HNO 2 HNO2 i NO 2 formed 3 3 6.000 10 mol 1.125 10 mol 3 4.875 10 mol Uit 4 Part B MHR 267
Determie the fial volume of the solutio. Use c to determie the cocetratios of HNO 2 (aq) ad NO 2 (aq) i mol/l. Write equatios for ay equilibrium reactios occurrig ad determie whether the solutio has bufferig capacity. Write the equilibrium expressio, substitute i the calculated cocetratios ad solve for the cocetratio of the hydroium io. HNO2 KOH 50.00 ml + 11.25 ml 1 L = 61.25 ml 1000 ml 0.06125 L chno 2 3 4.87510 mol 0.06125 L 2 7.959210 mol/l c NO 2 3 1.12510 mol 0.06125 L 2 1.836710 mol/l HNO 2 (aq) + H 2 O(l) NO 2 (aq) + H 3 O + (aq) NO 2 (aq) + H 2 O(l) HNO 2 (aq) + OH (aq) The solutio has bufferig capacity. The reactios of HNO 2 ad NO 2 reactig with water balace each other ad very little chage actually occurs. Therefore, you ca use the amouts of HNO 2 ad NO 2 preset after the reactio with KOH ad the equilibrium costat to fid the cocetratio of hydroium io, H 3 O + preset. [NO 2 ][H3O ] Ka [HNO 2 ] 2 4 (1.836710 )( x) 5.6 10 2 7.959210 x 1.836710 3 2.43267 10 4 2 (5.6 10 )(7.9592 10 ) 2 [H 3 O + ] = x = 2.43267 10-3 mol/l 268 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
Calculate ph usig: ph = log [H 3 O + ]. ph = log [H 3 O + ] = log (2.43267 10-3 ) = 2.615 = 2.61 Check Your Solutio The ph is cosistet with havig ureacted HNO 2 preset. The umber of sigificat digits to the right of the decimal is the same as the umber of sigificat digits i the give K a. The aswer seems reasoable. Uit 4 Part B MHR 269
106. If 100.00 ml of a 0.400 mol/l hydrofluoric acid solutio, HF(aq), is titrated with a 0.2000 mol/l sodium hydroxide solutio, (aq), determie the ph that results whe 20.00 ml of the base is added. For the hydrofluoric acid solutio, K a = 6.3 10 4. What Is Required? You eed to fid the ph at a certai poit i the titratio of hydrofluoric acid with sodium hydroxide solutio. What Is Give? You kow the volume of the hydrofluoric acid: HF 100.00 ml You kow the cocetratio of the hydrofluoric acid: chf 0.4000 mol/l You kow the volume of the sodium hydroxide: 20.00 ml You kow the cocetratio of the sodium hydroxide: c 0.2000 mol/l You are give the acid dissociatio costat for hydrofluoric acid: K a = 6.3 10 4 Pla Your Strategy Write the balaced equatio for the reactio betwee HF (aq) ad (aq). Use the formula c to calculate the amout i moles,, of HF(aq) preset before ay base was added. Use the formula c to calculate the amout i moles,, of (aq). Calculate the amout i moles,, of F (aq) preset after the base was added. Calculate the amout i moles of ureacted HF(aq). Act o Your Strategy Because is a strog base, the reactio essetially goes to completio. HF(aq) (aq) + (aq) NaF(aq) + H 2 O(l) HF c mol 0.4000 100.00 ml L 2 4.00010 mol c 1000 ml 1 L 0.2000 mol/ L 20.00 ml 1 L 1000 ml 4.000 10 mol Because the reactio goes to completio ad because the ratio of F to is 1:1, for every mole of added, oe mole of HF will be coverted ito F. Therefore: F 3 4.000 10 mol excess HF HF i F formed 2 3 4.000 10 mol 4.000 10 mol 2 3.600 10 mol 270 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
Determie the fial volume of the solutio. Use c to determie the cocetratios of HF(aq) ad F (aq) i mol/l. Write equatios for ay equilibrium reactios occurrig ad determie whether the solutio has bufferig capacity. Write the equilibrium expressio, substitute i the calculated cocetratios ad solve for the cocetratio of the hydroium io. HF 100.00 ml + 20.00 ml 1 L = 120.00 ml 1000 ml 0.1200 L chf 2 3.60010 mol 0.1200 L 1 0.30010 mol/l c F 3 4.00010 mol 0.1200 L 2 3.33310 mol/l HF(aq) + H 2 O(l) F (aq) + H 3 O + (aq) F (aq) + H 2 O(l) HF(aq) + OH (aq) The solutio has bufferig capacity. The reactios of HF ad F reactig with water balace each other ad very little chage actually occurs. Therefore, you ca use the amouts of HF ad F preset after the reactio with ad the equilibrium costat to fid the cocetratio of hydroium io, H 3 O + preset. [F ][H3O ] Ka [HF] 2 4 (3.33310 )( x) 6.3 10 1 3.00010 x 3.33310 5.6706 10 4 1 (6.3 10 )(3.000 10 ) 2 [H 3 O + ] = x = 5.6706 10 mol/l Uit 4 Part B MHR 271
Calculate ph usig: ph = log [H 3 O + ]. ph = log [H 3 O + ] = log (5.6706 10 ) = 2.2464 = 2.25 Check Your Solutio The ph is cosistet with havig ureacted HF preset. The umber of sigificat digits to the right of the decimal is the same as the umber of sigificat digits i the give K a. The aswer seems reasoable. 272 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4