Question #ab9ff

1 Answer
Dec 10, 2016

Answer:

Warning! Very long Answer! #"pH = (a) 9.26"; "(b) 5.23"; "(c) 4.51"; "(d) 3.13"#.

Explanation:

(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#