# Question ab9ff

Dec 10, 2016

Warning! Very long Answer! $\text{pH = (a) 9.26"; "(b) 5.23"; "(c) 4.51"; "(d) 3.13}$.

#### Explanation:

(a) pH at the beginning

We have an aqueous solution of the weak base, pyridine.

We set up an ICE table as usual.

$\textcolor{w h i t e}{m m m m m m l} \text{C"_5"H"_5"N" + "H"_2"O" ⇌ "C"_5"H"_5"N"^"+" + "OH"^"-}$
$\textcolor{w h i t e}{m m m m m m m l l} \text{Py" color(white)(ll)+ "H"_2"O" ⇌ color(white)(m)"PyH"^"+" color(white)(l)+ "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.190 \textcolor{w h i t e}{m m m m m m m l l} 0 \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mml)"-"x} \textcolor{w h i t e}{m m m m m m m l} + x \textcolor{w h i t e}{m m l} + x$
$\text{E/mol·L"^"-1":color(white)(ml)"0.190-"x} \textcolor{w h i t e}{m m m m m m m} x \textcolor{w h i t e}{m m m m} x$

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

$\text{pH" = "14.00 - pH" = "14.00 - 4.74} = 9.26$

(b) After adding 12.5 mL of acid

$\text{Initial moles of Py" = 25.0 color(red)(cancel(color(black)("mL Py"))) × "0.190 mmolL Py"/(1 color(red)(cancel(color(black)("mL Py")))) = "4.750 mmol Py}$

$\text{Moles of HBr added" = 12.5 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "2.375 mmol}$

$\textcolor{w h i t e}{m m m m m m} \text{Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m m} 4.750 \textcolor{w h i t e}{m l} 2.375 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol":color(white)(ml)"-2.375"color(white)(l)"+2.375} \textcolor{w h i t e}{m m l l} + 2.375$
$\text{E/mol} : \textcolor{w h i t e}{m l l} 2.375 \textcolor{w h i t e}{m m l} 0 \textcolor{w h i t e}{m m m m m l} 2.375$

The solution contains 2.375 mmol of $\text{Py}$ and 2.375 mmol of $\text{PyH"^"+}$.

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.

$\text{pH" = "14.00 - pOH" = "14.00 - 8.77} = 5.23$

(c) After adding 21 mL of acid

$\text{ Moles of HBr added" =21 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "3.99 mmol H"_3"O"^"+}$

$\textcolor{w h i t e}{m m m m m m} \text{Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m m} 4.750 \textcolor{w h i t e}{m l l} 3.99 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol":color(white)(ml)"-3.99"color(white)(mm)"-3.99"color(white)(mmll)"+3.99}$
$\text{E/mol} : \textcolor{w h i t e}{m l l} 0.76 \textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m m m l l} 3.99$

$\text{pOH" = "p"K_"b" + log((["PyH"^"+"])/(["Py"])) = -log(1.7× 10^"-9") + log((3.99 color(red)(cancel(color(black)("mmol"))))/(0.76 color(red)(cancel(color(black)("mmol"))))) = "8.77 + 0.72} = 9.49 .$

$\text{pH" = "14.00 -pOH" = "14.00 - 9.49} = 4.51$

(d) After adding 25 mL of acid

$\text{ Moles of HBr added" =25 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "4.75 mmol H"_3"O"^"+}$

$\textcolor{w h i t e}{m m m m m m} \text{Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m m} 4.750 \textcolor{w h i t e}{m m} 4.75 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol":color(white)(ml)"-4.75"color(white)(mm)"-4.75"color(white)(mmll)"+4.75}$
$\text{E/mol} : \textcolor{w h i t e}{m m l l} 0 \textcolor{w h i t e}{m m m l} 0 \textcolor{w h i t e}{m m m m l l} 4.75$

We have a solution that contains 4.75 mmol of $\text{PyH"^"+}$.

$\text{Volume" = "25.0 mL + 25 mL" = "50 mL}$

["PyH"^"+"] = "4.75 mmol"/"50 mL" = "0.095 mol/L"

$\textcolor{w h i t e}{m m m m m l} \text{PyH"^"+" + "H"_2"O" ⇌ color(white)(m)"Py"color(white)(l)+ "H"_3"O"^"+}$
$\text{I/mol} : \textcolor{w h i t e}{m m} 0.095 \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol":color(white)(mml)"-"xcolor(white)(mmmmmmmll)"+"xcolor(white)(mm)"+} x$
$\text{E/mol":color(white)(ml)"0.095-} x \textcolor{w h i t e}{m m m m m m l l} x \textcolor{w h i t e}{m m m} x$

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#