# A 35.0mL sample of 0.150 M acetic is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added?

## a) 0 mL b) 17.5 mL c) 34.5 mL d) 35.0 mL e) 35.5 mL f) 50.0 mL Thank you so much!

Feb 17, 2017

WARNING! Very long answer! The pH values are a) 2.79; b) 4.75; c) 6.59; d) 8.81;
e) 11.03; f)12.42.

#### Explanation:

a) 0 mL

$\textcolor{w h i t e}{m m m m m m m l} \text{HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.150 \textcolor{w h i t e}{m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmll)"-"xcolor(white)(mmmmml)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{l} 0.150 - x \textcolor{w h i t e}{m m m m l l} x \textcolor{w h i t e}{m m m} x$

K_"a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = x^2/(0.150-x) = 1.76 × 10^-5

0.150/(1.76 × 10^"-5") = 8523 > 400. ∴ x ≪ 0.150

x^2 = 0.150 × 1.76 × 10^"-5" = 2.64 × 10^"-6"

x = sqrt(2.64 × 10^"-6") = 1.62 × 10^"-3"

["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 1.62 × 10^"-3"color(white)(l) "mol/L"

"pH" = -log["H"_3"O"^"+"] = -log(1.62 × 10^"-3") = 2.79

b) At 17.5 mL

This is the point of half-neutralization.

;:"pH" = pK_"a" = -log(1.76 × 10^"-5") = 4.75

c) At 34.5 mL

$\textcolor{w h i t e}{m m m m m m} \text{HA" +color(white)(m) "OH"^"-"color(white)(m) ⇌ color(white)(m)"A"^"-" + "H"_2"O}$
$\text{I/mmol} : \textcolor{w h i t e}{m l} 5.25 \textcolor{w h i t e}{m m l} 5.175 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mmol":color(white)(ll)"-5.175"color(white)(ml)"-5.175"color(white)(mm)"+5.175}$
$\text{E/mmol} : \textcolor{w h i t e}{m} 0.075 \textcolor{w h i t e}{m m} 0 \textcolor{w h i t e}{m m m m l l} 5.175$

"Initial moles of HA" = 0.0350 color(red)(cancel(color(black)("L"))) × "0.150 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 25 mol" = color(red)(bb"5.25 mmol")

$\text{Moles of NaOH added" = 0.0345 color(red)(cancel(color(black)("L"))) × "0.150 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 175 mol" = "5.175 mmol}$

$\text{Moles of HA remaining" = ("5.25 -5.175) mmol" = "0.075 mmol}$

"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"])) = 4.75 + log((5.175 color(red)(cancel(color(black)("mol"))))/(0.075 color(red)(cancel(color(black)("mol"))))) = 4.75 + log(69.0) = 4.75 + 1.84 = 6.59

d) At 35.0 mL

$\textcolor{w h i t e}{m m m m m m} \text{HA" +color(white)(m) "OH"^"-"color(white)(m) ⇌ color(white)(m)"A"^"-" + "H"_2"O}$
$\text{I/mmol} : \textcolor{w h i t e}{m l} 5.25 \textcolor{w h i t e}{m m l} 5.25 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mmol":color(white)(ll)"-5.25"color(white)(mm)"-5.25"color(white)(mlmm)"+5.25}$
$\text{E/mmol} : \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m m m m} 0 \textcolor{w h i t e}{m m m m m l l} 5.25$

You have a solution of 5.25 mmol of $\text{A"^"-}$ in 70.0 mL.

["A"^"-"] = "5.25 mmol"/"70.0 mL" = "0.0750 mol/L"

$\textcolor{w h i t e}{m m m m m m m} \text{A"^"-" +color(white)(m) "H"_2"O"color(white)(m) ⇌ color(white)(m)"HA" + "OH"^"-}$
$\text{I/mmmol} : \textcolor{w h i t e}{m m l} 0.0750 \textcolor{w h i t e}{m m m m m m m m l l} 0 \textcolor{w h i t e}{m m l} 0$
$\text{C/mmol":color(white)(mmmm)"-"xcolor(white)(mmmmmmmmm)"+"xcolor(white)(mll)"+} x$
$\text{E/mmol} : \textcolor{w h i t e}{m m l} 0.0750 - x \textcolor{w h i t e}{m m m m m m m l} x \textcolor{w h i t e}{m m l} x$

${K}_{\text{b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(1.76 × 10^"-5") = 5.68 × 10^"-10}}$

${K}_{\text{b" = (["HA"]["OH"^"-"])/(["A"^"-"]) = x^2/(0.0750-x) = 5.68 × 10^"-10}}$

0.0750/(5.68×10^"-10") = 1.32 × 10^8 > 400. ∴ x ≪ 0.0750

x^2 = 0.0750 × 5.68 × 10^"-10" = 4.26 × 10^"-11"

x = sqrt(4.26 × 10^"-11") = 6.53 × 10^"-6"

["OH"^"-"] = 6.53 × 10^"-6" color(white)(l)"mol/L"

"pOH" = -log(6.53 × 10^"-6") = 5.19

$\text{pH" = "14.00 - pOH} = 14.00 - 5.19 = 8.81$

e) At 35.5 mL

You are now adding excess moles of $\text{OH"^"-}$

$\text{Excess moles of OH"^"-" = 0.5 color(red)(cancel(color(black)("mL OH"^"-"))) × "0.150 mmol OH"^"-"/(1 color(red)(cancel(color(black)("mL OH"^"-")))) = "0.075 mmol OH"^"-}$

["OH"^"-"] = "0.075 mmol"/"70.5 mL" = 1.06 × 10^"-3" "mol/L"

"pOH" = -log(1.06 × 10^"-3" ) = 2.97

$\text{pH" = "14.00 - 2.97} = 11.03$

f) At 50.00 mL

$\text{Excess moles of OH"^"-" = 15.0 color(red)(cancel(color(black)("mL OH"^"-"))) × "0.150 mmol OH"^"-"/(1 color(red)(cancel(color(black)("mL OH"^"-")))) = "2.25 mmol OH"^"-}$

["OH"^"-"] = "2.25 mmol"/"85.0 mL" = "0.0265 mol/L"#

$\text{pOH} = - \log \left(0.0265\right) = 1.58$

$\text{pH" = "14.00 - 1.58} = 12.42$

Your calculated values should match the titration curve below.