Question

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 `"HClO"; K_text(a) = 2.9 × 10^"-8"`
0.10 mol/L `"HClO"_3`
0.10 mol/L `"LiClO"_4`
0.10 mol/L `"CuSO"_4`

Answer 1

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 `"HClO"`

`"HClO"` is a weak acid with `K_"a" = 2.9 × 10^"-8"`

`color(white)(mmmmmmm)"HClO" + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"^"-"`
`"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmmmmm)0color(white)(mmmm)0`
`"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmm) "+"x`
`"E/mol·L"^"-1":color(white)(l)0.10 - xcolor(white)(mmmmmml) xcolor(white)(mmmll)x`

`K_"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"`

`"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 `"HClO"_3`

`"HClO"_3` is a strong acid.

`color(white)(mmmmmmll)"HClO"_3 + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"_3^"-"`
`"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmmmmml)0color(white)(mmmm)0`
`"C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mmmmmm)"+0.10"color(white)(mll) "+0.10"`
`"E/mol·L"^"-1":color(white)(mml)0color(white)(mmmmmmmll) 0.10color(white)(mmll)0.10`

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

`"Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L"`

0.10 mol/L `"LiClO"_4` and 0.10 mol/L `"CuSO"_4`

`"LiClO"_4` and `"CuSO"_4` are strong electrolytes.

Here is the calculation for `"LiClO"_4`.

`color(white)(mmmmmmm)"LiClO"_4 ⇌ "Li"^+ + "ClO"_4^"-"`
`"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmll)0color(white)(mmml)0`
`"C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mml)"+0.10"color(white)(m) "+0.10"`
`"E/mol·L"^"-1":color(white)(mml)0color(white)(mmmm) 0.10color(white)(mm)0.10`

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

`"Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L"`