ࡱ> z|y` .bjbj 1.H@H@H@H@$A?W@ CEEEEEEESSSSdSlTUlV$XhZ~VEEEEEVEEVnInInIE>EESnIESnInInIEC EA֜kH@0HnII W0?WnIe[He[nIe[nI EEnIEEEEEVVdI EEE?WEEEED<H@H@  Name: ________________________ Hour: ____ Date: ___________ Chemistry: pH and pOH calculations Part 1: Fill in the missing information in the table below. pH[ H3O1+ ]pOH[ OH1 ]ACID or BASE?3.783.89 x 104 M5.194.88 x 106 M8.468.45 x 1013 M2.142.31 x 1011 M10.917.49 x 106 M9.942.57 x 108 M4.161.06 x 101 M3.828.53 x 107 M7.054.73 x 1010 M1.339.87 x 103 M11.689.22 x 108 M12.245.39 x 1012 M Part 2: For each of the problems below, assume 100% dissociation. 1. A. Write the equation for the dissociation of hydrochloric acid. B. Find the pH of a 0.00476 M hydrochloric acid solution. 2. 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. 3. A. Write the equation for the dissociation of sodium hydroxide. B. Find the pH of a 0.000841 M solution of sodium hydroxide. 4. 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? 5. 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 in the solution? KEY Chemistry: pH and pOH calculations Part 1: Fill in the missing information in the table below. pH[ H3O1+ ]pOH[ OH1 ]ACID or BASE?3.781.66 x 104 M10.226.03 x 1011 MAcid3.413.89 x 104 M10.592.57 x 1011 MAcid8.811.55 x 109 M5.196.46 x 106 MBase8.692.04 x 109 M5.314.88 x 106 MBase8.463.47 x 109 M5.542.88 x 106 MBase12.18.45 x 1013 M1.901.26 x 102 MBase11.861.38 x 1012 M2.147.24 x 103 MBase3.403.98 x 104 M10.62.31 x 1011 MAcid10.911.23 x 1011 M3.098.13 x 104 MBase5.137.49 x 106 M8.871.35 x 109 MAcid4.068.71 x 105 M9.941.15 x 1010 MAcid6.413.89 x 107 M7.592.57 x 108 MAcid4.166.92 x 105 M9.841.45 x 1010 MAcid0.981.06 x 101 M13.01.00 x 1013 MAcid10.186.61 x 1011 M3.821.51 x 104 MBase7.931.17 x 108 M6.078.53 x 107 MBase7.058.91 x 108 M6.951.12 x 107 M~Base9.334.73 x 1010 M4.672.14 x 105 MBase12.672.14 x 1013 M1.334.68 x 102 MBase12.01.0 x 1012 M2.019.87 x 103 MBase11.682.09 x 1012 M2.324.79 x 103 MBase7.049.22 x 108 M6.961.10 x 107 M~Base1.761.74 x 102 M12.245.75 x 1013 MAcid2.702.00 x 103 M11.35.39 x 1012 MAcid Part 2: For each of the problems below, assume 100% dissociation. 1. A. Write the equation for the dissociation of hydrochloric acid. HCl(aq) ( H1+(aq) + Cl1-(aq) B. Find the pH of a 0.00476 M hydrochloric acid solution. HCl(aq) ( H1+(aq) + Cl1-(aq) 0.00476 M 0.00476 M  EMBED Equation.3  2. A. Write the equation for the dissociation of sulfuric acid. H2SO4(aq) ( 2 H1+(aq) + SO42-(aq) B. Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.  EMBED Equation.3  3. A. Write the equation for the dissociation of sodium hydroxide. NaOH(aq) ( Na1+(aq) + OH1-(aq) 0.000841 M 0.000841 M B. Find the pH of a 0.000841 M solution of sodium hydroxide.  EMBED Equation.3   EMBED Equation.3  or  EMBED Equation.3   EMBED Equation.3  Part 2: continued 4. A. Write the equation for the dissociation of aluminum hydroxide. Al(OH)3(aq) ( Al3+(aq) + 3 OH1-(aq) 2.36x10-5 M 7.08x10-5 M B. If the pH is 9.85, what is the concentration of the aluminum hydroxide solution?  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  2nd log  EMBED Equation.3   EMBED Equation.3  5. A. Write the equation for the dissociation of calcium hydroxide. Ca(OH)2(aq) ( Ca2+(aq) + 2 OH1-(aq) 2.18x10-3 M 4.37x10-5 M B. If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution?  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  EMBED Equation.3  2nd log  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  1a.  EMBED Equation.3 HCl(aq) ( H1+(aq) + Cl1-(aq) pH and pOH b. HCl(aq) ( H1+(aq) + Cl1-(aq) 0.00476 M 0.00476 M  EMBED Equation.3  2a.  EMBED Equation.3  3. NaOH(aq) ( Na1+(aq) + OH1-(aq) 0.000841 M 0.000841 M  EMBED Equation.3   EMBED Equation.3  or  EMBED Equation.3   EMBED Equation.3  4. Al(OH)3(aq) ( Al3+(aq) + 3 OH1-(aq) 2.36x10-5 M 7.08x10-5 M  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  2nd log  EMBED Equation.3   EMBED Equation.3  5. 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M=0.033 mol H 2 SO 4 2.75 L!  M = 0.0121 M H 2 SO 4 Step 3:  H 2 SO 4 (aq)   !    2 H +     +    SO 4  2" (aq)             0.0121 M         4"0.0242MStep 4:  pH  =  -log [H + ]  !  pH  =  -log [0.0242 M]  !  pH  =  1.62 FMicrosoft Equation 3.0 DS Equation Equation.39qH pOH  =CompObj$fObjInfo&Equation Native '_1136457346F>֜k>֜k  -log [OH - ]pOH  =  -log [0.000841 M]pOH  =  3.08 FMicrosoft Equation 3.0 DS Equation Equation.39qOle ,CompObj-fObjInfo/Equation Native 0"D pH  +  pOH  =  14pH  +3.08=14 pH  =  10.92 FMicrosoft Equation 3.0 DS Equation Equation.39q_11366985951F>֜k>֜kOle 4CompObj5fObjInfo7Equation Native 8_1136780397v]F>֜k>֜kOle >CompObj ?fHcpQ+ K w   =  [H + ][OH - ]110 -14   =  [H + ][0.000841 M][H + ]  = [1.1910 -11  M]  FMicrosoft Equation 3.0 DS Equation Equation.39qfT pH  =  -log[H + ]pH  =  -log [1.1910 -11  M]pH  =  10.92ObjInfo!AEquation Native B_1136457794{$F>֜k>֜kOle GCompObj#%HfObjInfo&JEquation Native K_1136698696E@)F>֜k>֜k FMicrosoft Equation 3.0 DS Equation Equation.39q" pH  +  pOH  =  149.85  + pOH=14 pOH  =  4.15Ole OCompObj(*PfObjInfo+REquation Native S FMicrosoft Equation 3.0 DS Equation Equation.39qH` j pOH  =  -log [OH - ]4.15  =  -log [OH - ]_1136457882.F>֜k>֜kOle WCompObj-/XfObjInfo0Z FMicrosoft Equation 3.0 DS Equation Equation.39q" | 7.08 x 10 -5  M3=  2.36 x 10 -5  M FMicrosoft Equation 3.0 DS EqEquation Native [_11366986943F>֜k>֜kOle ^CompObj24_fuation Equation.39qHGXV4O -4.15 = [OH - ] FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo5aEquation Native bc_11366987618F>֜k>֜kOle dCompObj79efObjInfo:gEquation Native h_1136459443qI=F>֜k>֜kHqHl [OH - ]  =  7.0810 -5  M FMicrosoft Equation 3.0 DS Equation Equation.39q"C pH  + Ole kCompObj<>lfObjInfo?nEquation Native o pOH  =  1411.64  + pOH=14 pOH  =  2.36 FMicrosoft Equation 3.0 DS Equation Equation.39qH4W pOH  =_11366988046SBF>֜k>֜kOle sCompObjACtfObjInfoDvEquation Native w_1136459757;GF>֜k>֜kOle {ObjInfoFH|  -logOH - []2.36  =  -log [OH - ]"~x w  2(4.3710 "3 M2.1810 -3  MEquation Native }_1136459735KF>֜k>֜kOle CompObjJLf FMicrosoft Equation 3.0 DS Equation Equation.39q"p  FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoMEquation Native #_1136698817PFEA֜kEA֜kOle CompObjOQfObjInfoREquation Native __1136698838NUFEA֜kEA֜kHCHQlO -2.36 =[OH - ] FMicrosoft Equation 3.0 DS Equation Equation.39qHqH , [OH - ]Ole CompObjTVfObjInfoWEquation Native   =  4.3710 -3  M FMicrosoft Equation 3.0 DS Equation Equation.39qv Rl M = molL!  2.1810 -_1207802323ZF>֜k>֜kOle CompObjY[fObjInfo\Equation Native _1207802301 X_F>֜k>֜kOle CompObj^`f3  M    !   x mol Ca(OH) 2 2.55 L !  x = 5.5710 -3 mol Ca(OH) 2 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoaEquation Native _1136459423dFEA֜kEA֜kOle n x g Ca(OH) 2  =  5.5710 -3  mol74 g Ca(OH) 2 1 mol Ca(OH) 2 ()=  0.412 g Ca(OH) 2 FMicrosoft Equation 3.0 DS Equation Equation.39q"  FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjcefObjInfofEquation Native #_1136392637iF>֜k>֜kOle CompObjhjfObjInfokEquation Native #l8.,  FMicrosoft Equation 3.0 DS Equation Equation.39qHgd Step 1:  x mol H 2 SO_1136698529nF>֜k>֜kOle CompObjmofObjInfopEquation Native _1136457161g"sF>֜k>֜kOle CompObjrtf 4 = 325 g H 2 SO 4 1 mol H 2 SO 4 98 g H 2 SO 4 ()=0.03 mol H 2 SO 4 Step 2:  M=molL!  M=0.03 mol H 2 SO 4 2.75 L!  M = 0.01 M H 2 SO 4 Step 3:  H 2 SO 4 (aq)   !    2 H +     +    SO 4  2" (aq)             0.01 M            4"0.02 MStep 4:  pH  =  -log [H + ]  !  pH  =  -log [0.02 M]  !  pH  =  1.70 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfouEquation Native _1136699046xFEA֜kEA֜kOle "- pH  =  -logH + []pH  =  -log1.1910 -11  M[]pH  =  10.92 FMicrosoft Equation 3.0 DS EqCompObjwyfObjInfozEquation Native ~_1136459290,b}FEA֜kEA֜kuation Equation.39qHbXV4E M = molL!  2.1810 -3  M    !   x mol Ca(OH) 2 L !  x = 5.5710 -3 mol Ca(OH) 2Ole CompObj|~fObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q"f- x g Ca(OH) 2  =  5.5710 -3  M74 g Ca(OH) 2 1 mol Ca(OH) 2 ()=  0.412 g Ca(OH) 2Oh+'0  ( H T ` lx pH & pOH ChemistryJeff ChristophersonNormalDd J  C A? 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