a) After adding 0.10 mL #"HCl"#
#"Initial moles NH"_3 = 0.010 color(red)(cancel(color(black)("L NH"_3))) × "0.10 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.0010 mol NH"_3#
#"Moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed"#
#= "0.000 10" color(red)(cancel(color(black)("L"))) × "1.00 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.000 10 mol"#
#"Moles NH"_3color(white)(l) "remaining = (0.0010 - 0.000 10) mol NH"_3 = "0.0009 mol NH"_3#
#"NH"_3 + "H"_2"O" ⇌ "NH"_4^+ "+ OH"^"-"; "pK"_text(b)" = 4.75"#
#"B + H"_2"O" ⇌ "BH"^+ "+ OH"^"-"#
#"pOH" = "pK"_"b" + log((["BH"^+])/(["B"]))#
#"pOH" = 4.75 + log(("0.000 10")/0.0009) = 4.75 - 0.95 = 3.80#
#"pH = 14.00 - pOH = 14.00 - 3.80 = 10.20"#
b) After adding another 0.10 mol #"HCl"#
#"Total moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed" = "0.000 20 mol"#
#"Moles NH"_3 "remaining = (0.0010 - 0.000 20) mol NH"_3 = "0.0008 mol NH"_3#
#"pOH" = 4.75 + log((["BH"^+])/(["B"])) = 4.75 + log(0.0002/0.0008) = 4.75 - 0.60 = 4.15#
#"pH = 14.00 - pOH = 14.00 - 4.15 = 9.85"#
The #"pH"# of this solution should be close to that in Part a), because the solution in Part a) is a buffer.
c) After adding 0.10 mL NaOH
The strong base will dissociate completely.
#"Moles of NaOH" = "0.000 10" color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = 1.0 × 10^"-5"color(white)(l) "mol"#
#V "= (10 + 0.10) mL = 10.1 mL = 0.0101 L"#
#["B"]_0 = "0.0010 mol"/"0.0101 L" = "0.099 mol/L"#
#["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0101 L") = 9.9 × 10^"-4"color(white)(l) "mol/L"#
#color(white)(mmmmmmmm)"B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mm) 0.099color(white)(mmmmml)0 color(white)(mmll)9.9 × 10^"-4"#
#"C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+"x#
#"E/mol·L"^"-1":color(white)(ml)"0.099 -" x color(white)(mmmmll)xcolor(white)(ml)9.9 × 10^"-4" + x#
#K_"b" = (x(1.0 × 10^"-5" + x))/("0.099 -" x) = 10^"-4.75" = 1.78 × 10^"-5"#
#x# is negligible with respect to #0.099#, but not with respect to #9.9 × 10^"-4"#.
#(x(1.0 × 10^"-5" + x))/0.099 = 1.78 × 10^"-5"#
#1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.099 = 1.76 × 10^"-8"#
#x^2 + 1.0 × 10^"-5"x - 1.76 × 10^"-8" = 0#
#x = 1.25 × 10^"-4"#
#["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"#
#"pOH = -log"(1.25 × 10^"-4") = 3.90#
#"pH = 14.00 - 3.90 = 10.10"#
d) After adding another 0.10 mL #"NaOH"#
#"Moles of NaOH added" = 0.000 20 color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH"))))#
#= 2.0 × 10^"-5"color(white)(l) "mol"#
#V "= (10 + 0.20) mL = 10.2 mL = 0.0102 L"#
#["B"]_0 = "0.0010 mol"/"0.0102 L" = "0.098 mol/L"#
#["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0102 L") = 9.8 × 10^"-4"color(white)(l) "mol/L"#
#color(white)(mmmmmmmm)"B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mm) 0.098color(white)(mmmmml)0 color(white)(mmll)9.8 × 10^"-4"#
#"C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+"x#
#"E/mol·L"^"-1":color(white)(ml)"0.098 -" x color(white)(mmmmll)xcolor(white)(ml)9.8 × 10^"-4" + x#
#K_"b" = (x(1.0 × 10^"-5" + x))/0.098 = 10^"-4.75" = 1.78 × 10^"-5"#
#x# is negligible with respect to 0.098, but not with respect to #9.8 × 10^"-4"#.
#(x(1.0 × 10^"-5" + x))/0.098 = 1.78 × 10^"-5"#
#1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.098 = 1.74 × 10^"-8"#
#x^2 + 1.0 × 10^"-5"x - 1.74 × 10^"-8" = 0#
#x = 1.25 × 10^"-4"#
#["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"#
#"pOH = -log"(1.25 × 10^"-4") = 3.90#
#"pH = 14.00 - 3.90 = 10.10"#
In this case, adding the extra #"NaOH"# has no effect because its concentration is negligible compared with that provided by the #"NH"_3#.
