# Calculate the pH after addition of (a) "30.5 mL" of "0.100 M HCl", and (b) at the second equivalence point, if the first equivalence point is reached after "25 mL" of "HCl" is added?

Aug 18, 2017

WARNING! Long answer! (a) pH = 5.82; (b) pH = 3.92.

#### Explanation:

The titration takes place in two stages:

In Stage 1, you are neutralizing the $\text{CO"_3^"2-}$:

$\text{CO"_3^"2-" + "H"_3"O"^"+" → "HCO"_3^"-" + "H"_2"O}$

In Stage 2, you are neutralizing the $\text{HCO"_3^"-}$:

$\text{HCO"_3^"-" + "H"_3"O"^"+" → "H"_2"CO"_3 + "H"_2"O}$

(a) pH at 30.5 mL $\text{HCl}$

At this point, you have used 25 mL of $\text{HCl}$ to reach the first equivalence point and have added 5.5 mL of $\text{HCl}$ towards the second equivalence point.

$\textcolor{w h i t e}{m m m m m l l} \text{HCO"_3^"-" + "H"_3"O"^"+" → "H"_2"CO"_3 + "H"_2"O}$
$\text{I/mmol} : \textcolor{w h i t e}{m l l} 2.50 \textcolor{w h i t e}{m m} 0.550 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mmol":color(white)(ll)"-0.550"color(white)(ml)"-0.550"color(white)(mm)"+0.550}$
$\text{E/mmol} : \textcolor{w h i t e}{m l} 1.95 \textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m m m} 0.550$

$\text{Initial moles HCO"_3^"-" = 0.025 color(red)(cancel(color(black)("L HCl"))) × (0.100 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("L HCl")))) × "1 mol HCO"_3^"-"/(1 color(red)(cancel(color(black)("mol HCl")))) = "0.002 50 mol HCO"_3^"-" = "2.50 mmol HCO"_3^"-}$

$\text{Added moles HCl" = 0.0055 color(red)(cancel(color(black)("L HCl"))) × "0.100 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.000 550 mol HCl" = "0.550 mmol HCl}$

We have a buffer consisting of 1.95 mmol $\text{HCO"_3^"-}$ and 0.550 mol ${\text{H"_2"CO}}_{3}$.

According to the Henderson-Hasselbalch equation,

color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log(["HCO"_3^"-"]/["H"_2"CO"_3])color(white)(a/a)|)))" "

For the first ionization of ${\text{H"_2"CO}}_{3}$, K_text(a) = 4.27 × 10^"-7";"p"K_text(a) = 6.37.

$\text{pH" = 6.37 + log(0.550/1.95) = "6.37 + log0.282" = "6.37 - 0.550} = 5.82$

pH at second equivalence point

At this stage we have added 50 mL of $\text{HCl}$ and have converted all of the $\text{CO"_3^"2-}$ to ${\text{H"_2"CO}}_{3}$.

$\text{Amt. of H"_2"CO"_3 = "2.50 mmol}$

$\text{Total volume" = "(25 + 50) mL" = "75 mL}$

["H"_2"CO"_3] = "2.50 mmol"/"75 mL" = "0.0333 mol/L"

$\textcolor{w h i t e}{m m m m m m l} \text{H"_2"CO"_3 + "H"_2"O" ⇌ "HCO"_3^"-" + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.0333 \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mmm)"+"x}$
$\text{E/mol·L"^"-1":color(white)(ll)"0.0333-} x \textcolor{w h i t e}{m m m m m m l} x \textcolor{w h i t e}{m m m l l} x$

K_text(a) = (["H"_3"O"^"+"]["HCO"_3^"-"])/(["H"_2"CO"_3]) = x^2/(0.0033-x) = 4.27 × 10^"-7"

0.0333/(4.27 × 10^"-7") = 7.81 × 10^4. ∴ x ≪0.0500

Then,

x^2 = 0.0333 × 4.27 × 10^"-7" = 1.42 × 10^"-8"

x = 1.19 × 10^"-4"

["H"_3"O"^"+"] = 1.19 × 10^"-4"color(white)(l) "mol/L"

"pH" = -log["H"_3"O"^"+"] = -log(1.19 × 10^"-4") = 3.92