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

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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 $HClO; K_{a} = 2.9 \times 10^{-8}$ $"HClO"; K_text(a) = 2.9 × 10^"-8"$
0.10 mol/L ${HClO}_{3}$ $"HClO"_3$
0.10 mol/L ${LiClO}_{4}$ $"LiClO"_4$
0.10 mol/L ${CuSO}_{4}$ $"CuSO"_4$

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Answer 1 · 2016-10-16T00:02:23

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 $1.0 \times 10^{-7} l {mol/L H}_{3} O^{+}$ $1.0 × 10^"-7"color(white)(l) "mol/L H"_3"O"^+$ and $1.0 \times 10^{-7} l {mol/L HO}^{-}$ $1.0 × 10^"-7"color(white)(l) "mol/L HO"^"-"$ , for a total ion concentration of $2.0 \times 10^{-7} l mol/L$ $2.0 × 10^"-7"color(white)(l) "mol/L"$ .

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: $2.0 \times 10^{-7} l mol/L$ $2.0 × 10^"-7"color(white)(l) "mol/L"$ .

0.10 mol/L $HClO$ $"HClO"$

$HClO$ $"HClO"$ is a weak acid with $K_{a} = 2.9 \times 10^{-8}$ $K_"a" = 2.9 × 10^"-8"$

$m m m m m m m HClO + H_{2} O ⇌ H_{3} O^{+} + {ClO}^{-}$ $color(white)(mmmmmmm)"HClO" + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"^"-"$
${I/mol\cdotL}^{-1} : m m 0.10 m m m m m m m 0 m m m m 0$ $"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmmmmm)0color(white)(mmmm)0$
${C/mol\cdotL}^{-1} : m m - x m m m m m m m + x m m m + x$ $"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmm) "+"x$
${E/mol\cdotL}^{-1} : l 0.10 - x m m m m m m l x m m m l l x$ $"E/mol·L"^"-1":color(white)(l)0.10 - xcolor(white)(mmmmmml) xcolor(white)(mmmll)x$

$K_{a} = \frac{[H_{3} O^{+}] [{ClO}^{-}]}{[HClO]} = \frac{x \times x}{0.10 - x} = 2.9 \times 10^{-8}$ $K_"a" = (["H"_3"O"^+]["ClO"^"-"])/"[HClO]"= (x×x)/(0.10-x) = 2.9 × 10^"-8"$

$\frac{0.10}{2.9 \times 10^{-8}} = 3.4 \times 10^{6}$ $0.10/(2.9 × 10^"-8") = 3.4 × 10^6$ , so $x ≪ 0.10$ $x ≪ 0.10$

∴ $\frac{x^{2}}{0.10} = 2.9 \times 10^{-8}$ $x^2/0.10 = 2.9 × 10^"-8"$

$x^{2} = 0.10 \times 2.9 \times 10^{-8} = 2.9 \times 10^{-9}$ $x^2 = 0.10 × 2.9 × 10^"-8" = 2.9 × 10^"-9"$

$x = 5.4 \times 10^{-5}$ $x = 5.4 × 10^"-5"$

∴ $[H_{3} O^{+}] = [{ClO}^{-}] = 5.4 \times 10^{-5} l mol/L$ $["H"_3"O"^+] = ["ClO"^"-"] = 5.4 × 10^"-5"color(white)(l) "mol/L"$

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

0.10 mol/L ${HClO}_{3}$ $"HClO"_3$

${HClO}_{3}$ $"HClO"_3$ is a strong acid.

$m m m m m m l l {HClO}_{3} + H_{2} O ⇌ H_{3} O^{+} + {ClO}_{3}^{-}$ $color(white)(mmmmmmll)"HClO"_3 + "H"_2"O" ⇌ "H"_3"O"^+ + "ClO"_3^"-"$
${I/mol\cdotL}^{-1} : m m 0.10 m m m m m m m l 0 m m m m 0$ $"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmmmmml)0color(white)(mmmm)0$
${C/mol\cdotL}^{-1} : m l - 0.10 m m m m m m +0.10 m l l +0.10$ $"C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mmmmmm)"+0.10"color(white)(mll) "+0.10"$
${E/mol\cdotL}^{-1} : m m l 0 m m m m m m m l l 0.10 m m l l 0.10$ $"E/mol·L"^"-1":color(white)(mml)0color(white)(mmmmmmmll) 0.10color(white)(mmll)0.10$

$[H_{3} O^{+}] = [{ClO}_{3}^{-}] = 0.10 mol/L$ $["H"_3"O"^+] = ["ClO"_3^"-"] = "0.10 mol/L"$

$Total ion concentration = 2 \times 0.10 mol/L = 0.20 mol/L$ $"Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L"$

0.10 mol/L ${LiClO}_{4}$ $"LiClO"_4$ and 0.10 mol/L ${CuSO}_{4}$ $"CuSO"_4$

${LiClO}_{4}$ $"LiClO"_4$ and ${CuSO}_{4}$ $"CuSO"_4$ are strong electrolytes.

Here is the calculation for ${LiClO}_{4}$ $"LiClO"_4$ .

$m m m m m m m {LiClO}_{4} ⇌ {Li}^{+} + {ClO}_{4}^{-}$ $color(white)(mmmmmmm)"LiClO"_4 ⇌ "Li"^+ + "ClO"_4^"-"$
${I/mol\cdotL}^{-1} : m m 0.10 m m m l l 0 m m m l 0$ $"I/mol·L"^"-1":color(white)(mm)0.10color(white)(mmmll)0color(white)(mmml)0$
${C/mol\cdotL}^{-1} : m l - 0.10 m m l +0.10 m +0.10$ $"C/mol·L"^"-1":color(white)(ml)"-"0.10color(white)(mml)"+0.10"color(white)(m) "+0.10"$
${E/mol\cdotL}^{-1} : m m l 0 m m m m 0.10 m m 0.10$ $"E/mol·L"^"-1":color(white)(mml)0color(white)(mmmm) 0.10color(white)(mm)0.10$

$[{Li}^{+}] = [{ClO}_{4}^{-}] = 0.10 mol/L$ $["Li"^+] = ["ClO"_4^"-"] = "0.10 mol/L"$

$Total ion concentration = 2 \times 0.10 mol/L = 0.20 mol/L$ $"Total ion concentration" = "2 × 0.10 mol/L" = "0.20 mol/L"$

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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 $HClO; K_{a} = 2.9 \times 10^{-8}$ $"HClO"; K_text(a) = 2.9 × 10^"-8"$
0.10 mol/L ${HClO}_{3}$ $"HClO"_3$
0.10 mol/L ${LiClO}_{4}$ $"LiClO"_4$
0.10 mol/L ${CuSO}_{4}$ $"CuSO"_4$

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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;Ka=2.9×10-8"HClO"; K_text(a) = 2.9 × 10^"-8" 0.10 mol/L HClO3"HClO"_3 0.10 mol/L LiClO4"LiClO"_4 0.10 mol/L CuSO4"CuSO"_4

1 Answer

Explanation:

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0.1 mol/L propanol
0.1 mol/L sucrose
0.1 mol/L $HClO; K_{a} = 2.9 \times 10^{-8}$ $"HClO"; K_text(a) = 2.9 × 10^"-8"$
0.10 mol/L ${HClO}_{3}$ $"HClO"_3$
0.10 mol/L ${LiClO}_{4}$ $"LiClO"_4$
0.10 mol/L ${CuSO}_{4}$ $"CuSO"_4$