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#

1 Answer
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 #"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"#