PH And POH - Ms. Mogck's Classroom - Ms. Mogck's Classroom
Name:Hour: Date:Chemistry: pH and pOH calculationsPart 1: Fill in the missing information in the table below.pH[ H3O1 ]pOH[ OH1–]ACID or BASE?3.783.89 x 10–4M5.19–64.88 x 10 M8.468.45 x 10–13M2.14–112.31 x 10M10.917.49 x 10–6M9.94–82.57 x 10 M4.161.06 x 10–1M3.82–78.53 x 10 M7.054.73 x 10–10M1.33–39.87 x 10 M11.689.22 x 10–8M12.24–125.39 x 10M
Part 2: For each of the problems below, assume 100% dissociation.1.2.3.4.5.A.Write the equation for the dissociation of hydrochloric acid.B.Find the pH of a 0.00476 M hydrochloric acid solution.A.Write the equation for the dissociation of sulfuric acid.B.Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.A.Write the equation for the dissociation of sodium hydroxide.B.Find the pH of a 0.000841 M solution of sodium hydroxide.A.Write the equation for the dissociation of aluminum hydroxide.B.If the pH is 9.85, what is the concentration of the aluminum hydroxide solution?A.Write the equation for the dissociation of calcium hydroxide.B.If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are inthe solution?
KEYChemistry: pH and pOH calculationsPart 1: Fill in the missing information in the table below.1 pH[ H3O3.781.66 x 103.413.89 x 108.811.55 x 108.692.04 x 108.463.47 x 1012.18.45 x 1011.861.38 x 103.403.98 x 1010.911.23 x 105.137.49 x 104.068.71 x 106.413.89 x 104.166.92 x 100.981.06 x 1010.186.61 x 107.931.17 x 107.058.91 x 109.334.73 x 1012.672.14 x 1012.01.0 x 1011.682.09 x ��5–7–5[ OH1–]–11M10.226.03 x 10M10.592.57 x 10MACID or BASE?MAcidMAcid5.196.46 x 10 MBaseM5.314.88 x 10 MM5.542.88 x 10 MM1.901.26 x 10 MM2.147.24 x 10 MM10.62.31 x 10MAcidM3.098.13 x 10 MBaseM8.871.35 x 10 MM9.941.15 x 10MAcidM7.592.57 x 10 MAcidM9.841.45 x 10M13.01.00 x –10BaseBaseBaseBaseAcidMAcidMAcid3.821.51 x 10 MBaseM6.078.53 x 10 MM6.951.12 x 10 MM4.672.14 x 10 MM1.334.68 x 10 MM2.019.87 x 10 MM2.324.79 x 10 M7.049.22 x 10 M6.961.10 x 10 M1.761.74 x 10M12.245.75 x 102.702.00 x 10M11.35.39 x se BaseBaseBaseBaseBase BaseMAcidMAcid
Part 2: For each of the problems below, assume 100% dissociation.1.A.Write the equation for the dissociation of hydrochloric acid.1 1-HCl(aq) H (aq) Cl (aq)B.Find the pH of a 0.00476 M hydrochloric acid solution.1 1-HCl(aq) H (aq) Cl (aq)0.00476 M0.00476 MpHpHpH2.A.- log [H ]- log [0.00476 M]2.32Write the equation for the dissociation of sulfuric acid.1 2-H2SO4(aq) 2 H (aq) SO4 (aq)B.Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.Step 1 : x mol H 2 SO 4molLStep 2 : M3.25 g H 2 SO 40.033 mol H 2 SO 42.75 LMStep 3 : H 2 SO 4 (aq )2H0.0121 MStep 4 : pH3.A.SO 420.033 mol H 2 SO 4M 0.0121 M H 2 SO 4(aq )0.0242M- log [H ]pH- log [0.0242 M]pH1.62Write the equation for the dissociation of sodium hydroxide.NaOH(aq)1 Na (aq)0.000841 MB.1 mol H 2 SO 498 g H 2 SO 41- OH (aq)0.000841 MFind the pH of a 0.000841 M solution of sodium hydroxide.- log [OH- ]- log [0.000841 M]3.08pOHpOHpOHpHpHpHpOH 143.08 1410.92pH- log[H ]pHpH- log [1.19 10 -11 M]10.92orKw[H ][OH- ]1 10 -14[H ][H ][0.000841 M][1.19 10 -11 M]
Part 2: continued4.A.Write the equation for the dissociation of aluminum hydroxide.3 Al (aq)Al(OH)3(aq)-51- 3 OH (aq)7.08x10-5 M2.36x10 MB.If the pH is 9.85, what is the concentration of the aluminum hydroxide solution?pHpOH 149.85 pOH 14pOH4.15- log [OH- ]pOH-4.15nd2- log [OH ]A.7.08 10-5 MWrite the equation for the dissociation of calcium hydroxide.Ca(OH)2(aq)2.18x10-3 MB.2.36 x 10 - 5 Mlog - 4.15 [OH- ][OH- ]5.7.08 x 10 -5 M32 Ca (aq)1- 2 OH (aq)4.37x10-5 MIf the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are inthe solution?pHpOH 1411.64 pOH 14pOH2.36nd2molL2.36- log [OH- ]2.18 10 -3 M2 4.37 10 3 M4.37 10-3 M2.18 10 - 3 Mx g Ca(OH) 2- log OH-log - 2.36 [OH- ][OH- ]MpOH5.57 10 - 3 Mx mol Ca(OH) 2L74 g Ca(OH) 21mol Ca(OH) 2x 5.57 10 - 3 mol Ca(OH) 20.412 g Ca(OH) 2
1 1-HCl(aq) H (aq) Cl (aq)1a.1 pH and pOH1-b. HCl(aq) H (aq) Cl (aq)0.00476 M0.00476 MpHpHpHStep 1 : x mol H2SO 4Step 2 : MmolLM0.03 mol H2SO 42.75 L2a. Step 3 : H2SO 4 (aq )2H0.01 MStep 4 : pH3. NaOH(aq)- log [H ]1 Na (aq)pH- log HpHpH- log 1.19 10 -11 M10.92-5pH3 Al (aq)pHpOH 149.85 pOH 14pOH4.15pHpHpH- log [0.02 M]pOH 143.08 1410.921.702 Ca (aq)pOH- log [OH- ]4.15- log [OH- ]1 10 -14[H ][H ][0.000841 M][1.19 10 -11 M]7.08 x 10 -5 M32.36 x 10 - 5 Mlog - 4.15 [OH- ]7.08 10-5 M1- 2 OH (aq)4.37x10-5 MpOH- log OH-2.36- log [OH- ]nd22.18 10 -3 M2 4.37 10 3 Mlog - 2.36 [OH- ][OH- ]4.37 10-3 Mx mol Ca(OH) 2x 5.57 10 - 3 mol Ca(OH) 2L74 g Ca(OH) 25.57 10 - 3 M0.412 g Ca(OH) 21mol Ca(OH) 22.18 10 - 3 Mx g Ca(OH) 2or[H ][OH- ]7.08x10-5 M2.18x10 MpHpOH 1411.64 pOH 14pOH2.36Kw1-[OH- ]molLpH 3 OH (aq)ndM(aq )1-2-32 OH (aq)2.36x10 M5. Ca(OH)2(aq)M 0.01 M H2SO 40.000841 M- log [OH- ]- log [0.000841 M]3.084. Al(OH)3(aq)SO 40.03 mol H2SO 40.02 M0.000841 MpOHpOHpOH1 mol H2SO 498 g H2SO 4325 g H2SO 4- log [H ]- log [0.00476 M]2.32
KEY Chemistry: pH and pOH calculations Part 1: Fill in the missing information in the table below. pH [ H 3 O 1 ] pOH [ OH1–] ACID or BASE? 3.78 1.66 x 10–4 M 10.22 6.03 x 10–11 M Acid 3.41 3.89 x 10–4 M 10.59 2.57 x 10–11 M Acid 8.81 1.55 x 10–9 M 5.19 6.46 x 10–6 M Base 8.69 2.04 x 10–9 M 5.31 4.88 x 10–6 M Base 8.46 3.47 x 10–9 M 5.54 2.88 x 10–6 M Base
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