a) 0 mL #color(white)(mmmmmmml)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"# #K_"a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = x^2/(0.150-x) = 1.76 × 10^-5# #0.150/(1.76 × 10^"-5") = 8523 > 400#. ∴ #x ≪ 0.150## #x^2 = 0.150 × 1.76 × 10^"-5" = 2.64 × 10^"-6"# #x = sqrt(2.64 × 10^"-6") = 1.62 × 10^"-3"# #["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 1.62 × 10^"-3"color(white)(l) "mol/L"# #"pH" = -log["H"_3"O"^"+"] = -log(1.62 × 10^"-3") = 2.79# b) At 17.5 mL This is the point of half-neutralization. ;:#"pH" = pK_"a" = -log(1.76 × 10^"-5") = 4.75# c) At 34.5 mL #color(white)(mmmmmm)"HA" +color(white)(m) "OH"^"-"color(white)(m) ⇌ color(white)(m)"A"^"-" + "H"_2"O"# #"Initial moles of HA" = 0.0350 color(red)(cancel(color(black)("L"))) × "0.150 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 25 mol" = color(red)(bb"5.25 mmol")# #"Moles of NaOH added" = 0.0345 color(red)(cancel(color(black)("L"))) × "0.150 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 175 mol" = "5.175 mmol"# #"Moles of HA remaining" = ("5.25 -5.175) mmol" = "0.075 mmol"# #"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"])) = 4.75 + log((5.175 color(red)(cancel(color(black)("mol"))))/(0.075 color(red)(cancel(color(black)("mol"))))) = 4.75 + log(69.0) = 4.75 + 1.84 = 6.59# d) At 35.0 mL #color(white)(mmmmmm)"HA" +color(white)(m) "OH"^"-"color(white)(m) ⇌ color(white)(m)"A"^"-" + "H"_2"O"# You have a solution of 5.25 mmol of #"A"^"-"# in 70.0 mL. #["A"^"-"] = "5.25 mmol"/"70.0 mL" = "0.0750 mol/L"# #color(white)(mmmmmmm)"A"^"-" +color(white)(m) "H"_2"O"color(white)(m) ⇌ color(white)(m)"HA" + "OH"^"-"# #K_"b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(1.76 × 10^"-5") = 5.68 × 10^"-10"# #K_"b" = (["HA"]["OH"^"-"])/(["A"^"-"]) = x^2/(0.0750-x) = 5.68 × 10^"-10"# #0.0750/(5.68×10^"-10") = 1.32 × 10^8 > 400#. ∴ #x ≪ 0.0750# #x^2 = 0.0750 × 5.68 × 10^"-10" = 4.26 × 10^"-11"# #x = sqrt(4.26 × 10^"-11") = 6.53 × 10^"-6"# #["OH"^"-"] = 6.53 × 10^"-6" color(white)(l)"mol/L"# #"pOH" = -log(6.53 × 10^"-6") = 5.19# #"pH" = "14.00 - pOH" = 14.00 - 5.19 = 8.81# e) At 35.5 mL You are now adding excess moles of #"OH"^"-"# #"Excess moles of OH"^"-" = 0.5 color(red)(cancel(color(black)("mL OH"^"-"))) × "0.150 mmol OH"^"-"/(1 color(red)(cancel(color(black)("mL OH"^"-")))) = "0.075 mmol OH"^"-"# #["OH"^"-"] = "0.075 mmol"/"70.5 mL" = 1.06 × 10^"-3" "mol/L"# #"pOH" = -log(1.06 × 10^"-3" ) = 2.97# #"pH" = "14.00 - 2.97" = 11.03# f) At 50.00 mL #"Excess moles of OH"^"-" = 15.0 color(red)(cancel(color(black)("mL OH"^"-"))) × "0.150 mmol OH"^"-"/(1 color(red)(cancel(color(black)("mL OH"^"-")))) = "2.25 mmol OH"^"-"# #["OH"^"-"] = "2.25 mmol"/"85.0 mL" = "0.0265 mol/L"# #"pOH" = -log(0.0265) = 1.58# #"pH" = "14.00 - 1.58" = 12.42# Your calculated values should match the titration curve below.
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Calculate the pH when 76.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴).
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