(a) pH at the beginning
We have an aqueous solution of the weak base, pyridine.
We set up an ICE table as usual.
color(white)(mmmmmml)"C"_5"H"_5"N" + "H"_2"O" ⇌ "C"_5"H"_5"N"^"+" + "OH"^"-"
color(white)(mmmmmmmll)"Py" color(white)(ll)+ "H"_2"O" ⇌ color(white)(m)"PyH"^"+" color(white)(l)+ "OH"^"-"
"I/mol·L"^"-1":color(white)(mm)0.190color(white)(mmmmmmmll)0color(white)(mmmm)0
"C/mol·L"^"-1":color(white)(mml)"-"x"color(white)(mmmmmmml)+xcolor(white)(mml)+x
"E/mol·L"^"-1":color(white)(ml)"0.190-"x"color(white)(mmmmmmm)xcolor(white)(mmmm)x
K_b = (["PyH"^+]["OH"⁻])/"[Py]" = = (x·x)/(0.190-x) = x^2/(0.190-x) = 1.7 × 10^"-9"
0.190/(1.7 × 10^"-9") = 1.1 × 10^8 >400; ∴ x ≪ 0.190.
x^2/0.190 = 1.7 × 10^"-9"
x^2 = 0.190 × 1.7 × 10^"-9" = 3.23 × 10^"-10"
x = 1.80 × 10^"-5"
["OH"^"-"] = xcolor(white)(l) "mol/L" = 1.80 × 10^"-5"color(white)(l) "mol/L"
"pOH" = "-log"["OH"^"-"] = "-log"(1.80 × 10^"-5") = 4.74
"pH" = "14.00 - pH" = "14.00 - 4.74" = 9.26
(b) After adding 12.5 mL of acid
"Initial moles of Py" = 25.0 color(red)(cancel(color(black)("mL Py"))) × "0.190 mmolL Py"/(1 color(red)(cancel(color(black)("mL Py")))) = "4.750 mmol Py"
"Moles of HBr added" = 12.5 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "2.375 mmol"
color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"
"I/mol":color(white)(mm)4.750color(white)(ml)2.375color(white)(mmmml)0
"C/mol":color(white)(ml)"-2.375"color(white)(l)"+2.375"color(white)(mmll)+2.375
"E/mol":color(white)(mll)2.375color(white)(mml)0color(white)(mmmmml)2.375
The solution contains 2.375 mmol of "Py" and 2.375 mmol of "PyH"^"+".
We have a buffer.
"pOH" = "p"K_"b" + log((["PyH"^"+"])/(["Py"])) = -log(1.7× 10^"-9") + log(stackrelcolor(blue)(1)(color(red)(cancel(color(black)"2.375 mmol")))/(color(red)(cancel(color(black)("2.375 mmol"))))) = 8.77 + 0 = 8.77.
"pH" = "14.00 - pOH" = "14.00 - 8.77" = 5.23
(c) After adding 21 mL of acid
" Moles of HBr added" =21 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "3.99 mmol H"_3"O"^"+"
color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"
"I/mol":color(white)(mm)4.750color(white)(mll)3.99color(white)(mmmml)0
"C/mol":color(white)(ml)"-3.99"color(white)(mm)"-3.99"color(white)(mmll)"+3.99"
"E/mol":color(white)(mll)0.76color(white)(mmm)0color(white)(mmmmll)3.99
"pOH" = "p"K_"b" + log((["PyH"^"+"])/(["Py"])) = -log(1.7× 10^"-9") + log((3.99 color(red)(cancel(color(black)("mmol"))))/(0.76 color(red)(cancel(color(black)("mmol"))))) = "8.77 + 0.72"=9.49.
"pH" = "14.00 -pOH" = "14.00 - 9.49" = 4.51
(d) After adding 25 mL of acid
" Moles of HBr added" =25 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "4.75 mmol H"_3"O"^"+"
color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"
"I/mol":color(white)(mm)4.750color(white)(mm)4.75color(white)(mmmml)0
"C/mol":color(white)(ml)"-4.75"color(white)(mm)"-4.75"color(white)(mmll)"+4.75"
"E/mol":color(white)(mmll)0color(white)(mmml)0color(white)(mmmmll)4.75
We have a solution that contains 4.75 mmol of "PyH"^"+".
"Volume" = "25.0 mL + 25 mL" = "50 mL"
["PyH"^"+"] = "4.75 mmol"/"50 mL" = "0.095 mol/L"
color(white)(mmmmml)"PyH"^"+" + "H"_2"O" ⇌ color(white)(m)"Py"color(white)(l)+ "H"_3"O"^"+"
"I/mol":color(white)(mm)0.095color(white)(mmmmmmml)0color(white)(mmm)0
"C/mol":color(white)(mml)"-"xcolor(white)(mmmmmmmll)"+"xcolor(white)(mm)"+"x
"E/mol":color(white)(ml)"0.095-"xcolor(white)(mmmmmmll)xcolor(white)(mmm)x
K_a = (["Py"]["H"_3"O"^"+"])/(["PyH"^"+"]) = K_w/K_b = (1.00 × 10^"-14")/(1.7 × 10^"-9") = 5.88 × 10^"-6"
x^2/("0.095-"x) = 5.88 × 10^"-6"
0.095/(5.88 × 10^"-6") = 1.2 × 10^4 > 400; ∴ x ≪ 0.095
x^2/0.095 = 5.88 × 10^"-6"
x^2 = 0.095 × 5.88 × 10^"-6" = 5.59 × 10^"-7"
x = 7.48 × 10^"-4"
["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 7.48 × 10^"-4"color(white)(l) "mol/L"
"pH" = -log(7.48× 10^"-4") = 3.13
