(a) pH of aspirin solution
Let's write the chemical equation as
#color(white)(mmmmmmmml)"HA" color(white)(m)+color(white)(m) "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"A"^"-"#
#"I/mol·dm"^"-3":color(white)(mm)0.05color(white)(mmmmmmmml)0color(white)(mmmmmll)0#
#"C/mol·dm"^"-3":color(white)(mml)"-"xcolor(white)(mmmmmmmm)"+"xcolor(white)(mlmmml)"+"x#
#"E/mol·dm"^"-3":color(white)(m) "0.05 -"color(white)(l)xcolor(white)(mmmmmmml)xcolor(white)(mmxmmm)x#
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/("0.05 -"color(white)(l)x) = 3.27 × 10^"-4"#
Check for negligibility
#0.05/(3.27 × 10^"-4") = 153 < 400#
∴ #x# is not less than 5 % of the initial concentration of #["HA"]#.
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
#x^2/(0.05 -x) = 3.27 × 10^"-4"#
#x^2 = 3.27×10^"-4"(0.05-x)= 1.635 × 10^"-5" - 3.27 × 10^"-4"x#
#x^2 + 3.27 × 10^"-4"x -1.635 ×10^"-5" =0#
#x = 1.68 × 10^"-5"#
#["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 1.68 × 10^"-5"color(white)(l) "mol/L"#
#"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.68 × 10^"-5") = 4.77#
(b) #["H"_3"O"^"+"]# at pH 4
#["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 1.00 ×10^"-4"color(white)(l) "mol/L"#
(c) Concentration of #"A"^"-"# in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the #["A"^"-"]#.
#"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]"))#
#4.00 = -log(3.27 × 10^"-4") + log((["A"^"-"])/0.05) = 3.49 + log((["A"^"-"])/0.05)#
#log((["A"^"-"])/0.05) = "4.00 - 3.49" = 0.51#
#(["A"^"-"])/0.05 = 10^0.51 = 3.24#
#["A"^"-"] = 0.05 × 3.24 = 0.16#
The concentration of #"A"^"-"# in the buffer is 0.16 mol/L.
