# What are the total concentrations of ions in the following solutions?

## 0.1 mol/L propanol 0.1 mol/L sucrose 0.1 mol/L $\text{HClO"; K_text(a) = 2.9 × 10^"-8}$ 0.10 mol/L ${\text{HClO}}_{3}$ 0.10 mol/L ${\text{LiClO}}_{4}$ 0.10 mol/L ${\text{CuSO}}_{4}$

Oct 16, 2016

#### Answer:

Warning! Long answer! These are the concentrations of ions in 0.1 mol/L aqueous solutions.

#### Explanation:

The calculations start from the premise that water contains 1.0 × 10^"-7"color(white)(l) "mol/L H"_3"O"^+ and 1.0 × 10^"-7"color(white)(l) "mol/L HO"^"-", for a total ion concentration of 2.0 × 10^"-7"color(white)(l) "mol/L".

0.1 mol/L Propanol and 0.10 mol/L Sucrose

Propanol and sucrose are nonelectrolytes.

They do not generate ions, so the total ion concentration is the same as in pure water: 2.0 × 10^"-7"color(white)(l) "mol/L".

0.10 mol/L $\text{HClO}$

$\text{HClO}$ is a weak acid with ${K}_{\text{a" = 2.9 × 10^"-8}}$

$\textcolor{w h i t e}{m m m m m m m} \text{HClO" + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.10 \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmm) "+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{l} 0.10 - x \textcolor{w h i t e}{m m m m m m l} x \textcolor{w h i t e}{m m m l l} x$

${K}_{\text{a" = (["H"_3"O"^+]["ClO"^"-"])/"[HClO]"= (x×x)/(0.10-x) = 2.9 × 10^"-8}}$

0.10/(2.9 × 10^"-8") = 3.4 × 10^6, so x ≪ 0.10

x^2/0.10 = 2.9 × 10^"-8"

x^2 = 0.10 × 2.9 × 10^"-8" = 2.9 × 10^"-9"

x = 5.4 × 10^"-5"

["H"_3"O"^+] = ["ClO"^"-"] = 5.4 × 10^"-5"color(white)(l) "mol/L"

$\text{Total ion concentration" = 2 × 5.4 × 10^"-5"color(white)(l) "mol/L" = 1.1 × 10^"-4" color(white)(l)"mol/L}$

0.10 mol/L ${\text{HClO}}_{3}$

${\text{HClO}}_{3}$ is a strong acid.

$\textcolor{w h i t e}{m m m m m m l l} \text{HClO"_3 + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"_3^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.10 \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mmmmmm)"+0.10"color(white)(mll) "+0.10}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m l} 0 \textcolor{w h i t e}{m m m m m m m l l} 0.10 \textcolor{w h i t e}{m m l l} 0.10$

["H"_3"O"^+] = ["ClO"_3^"-"] = "0.10 mol/L"

$\text{Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L}$

0.10 mol/L ${\text{LiClO}}_{4}$ and 0.10 mol/L ${\text{CuSO}}_{4}$

${\text{LiClO}}_{4}$ and ${\text{CuSO}}_{4}$ are strong electrolytes.

Here is the calculation for ${\text{LiClO}}_{4}$.

$\textcolor{w h i t e}{m m m m m m m} \text{LiClO"_4 ⇌ "Li"^+ + "ClO"_4^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.10 \textcolor{w h i t e}{m m m l l} 0 \textcolor{w h i t e}{m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mml)"+0.10"color(white)(m) "+0.10}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m l} 0 \textcolor{w h i t e}{m m m m} 0.10 \textcolor{w h i t e}{m m} 0.10$

["Li"^+] = ["ClO"_4^"-"] = "0.10 mol/L"

$\text{Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L}$