A solution is 1.0 M in acetic acid. Pure water is added to double the volume of the solution. ?

a) Does the percent ionization increase, decrease or remain the
same?

b) When the acid comes to equilibrium after dilution, is the value
of the reaction quotient increased, decreased, or the same as it was prior to
dilution?

1 Answer
Sep 8, 2016

Answer:

WARNING! Long answer. The % ionization increases, but the reaction quotient #Q# remains the same as it was prior to dilution.

Explanation:

(a) Effect of dilution on % ionization

The formula for % ionization is

#color(blue)(|bar(ul(color(white)(a/a)"% Ionization" = (["H"^+]_"eq")/(["HA"]_0) × 100 %color(white)(a/a)|)))" "#

Our first task, then is to calculate #["H"^+]_"eq"#.

We start with an ICE table.

#color(white)(mmmmmml)"HA" ⇌ "H"^+ + "A"^"-"#
#"I/mol·L"^"-1": color(white)(m)0.50 color(white)(mmll) 0color(white)(mmll)0#
#"C/mol·L"^"-1": color(white)(m)"-"xcolor(white)(mll)+x color(white)(m)+x#
#"E/mol·L"^"-1":color(white)(l) "0.50 -"xcolor(white)(mll)xcolor(white)(mml)color(white)(l)x#

#K_a = ([H^+][A^"-"])/([HA]) = (x × x)/(1.0 -x) = 1.75 × 10^"-5"#

#[HA]_0/K_a = 1.0/(1.75 × 10^"-5") = "57 000" >400#, so #x ≪ 1.0#.

Then #x^2/1.0 = 1.75 × 10^"-5"#

#x^2 = 1.75 × 10^"-5"#

#x = sqrt(1.75 × 10^"-5") = 4.18 × 10^"-3"#

#["H"^+]_"eq" = xcolor(white)(l) "mol/L" = 4.18 × 10^"-3"color(white)(l) "mol/L"#

#"% Ionization =" (["H"^+]_"eq")/(["HA"]_0) × 100 % = (4.18 × 10^"-3" color(red)(cancel(color(black)("mol/L"))))/(1.0 color(red)(cancel(color(black)("mol/L")))) × 100 % = 0.42 %#

Diluted acid

The diluted acid is in twice the volume, so its concentration has been halved to 0.50 mol/L.

#color(white)(mmmmmml)"HA" ⇌ "H"^+ + "A"^"-"#
#"I/mol·L"^"-1": color(white)(m)0.50 color(white)(mmll) 0color(white)(mmll)0#
#"C/mol·L"^"-1": color(white)(m)"-"xcolor(white)(mll)+x color(white)(m)+x#
#"E/mol·L"^"-1":color(white)(l) "0.50 -"xcolor(white)(mll)xcolor(white)(mml)color(white)(l)x#

#K_a = (["H"^+]["A"^"-"])/(["HA"]) = (x × x)/("1.0 -x") = 1.75 × 10^"-5"#

#["HA"]_0/K_a = 0.50/(1.75 × 10^"-5") = "29 000" >400#, so #x ≪ 0.50#.

Then #x^2/0.50 = 1.75 × 10^"-5"#

#x^2 = 0.50 × 1.75 × 10^"-5" = 8.75 × 10^"-6"#

#x = sqrt(8.75 × 10^"-6") = 2.96 × 10^"-3"#

#["H"^+]_"eq" = xcolor(white)(l) "mol/L" = 2.96 × 10^"-3"color(white)(l) "mol/L"#

#"% Ionization =" (["H"^+]_"eq")/(["HA"]_0) × 100 % = (2.96 × 10^"-3" color(red)(cancel(color(black)("mol/L"))))/(0.50 color(red)(cancel(color(black)("mol/L")))) × 100 % = 0.59 %#
∴ The % ionization increases as the concentration decreases.

(b) Effect of dilution on #Q#

The formula for the reaction quotient #Q# is

#color(blue)(|bar(ul(color(white)(a/a) Q = (["H"^+]["A"^"-"])/(["HA"])color(white)(a/a)|)))" "#

For the 1 mol/L acid,

#Q = (["H"^+]["A"^"-"])/(["HA"]) = (4.18 × 10^"-3" × 4.18 × 10^"-3")/1.0 = 1.75 × 10^"-5"#

For the 0.50 mol/L acid,

#Q = (["H"^+]["A"^"-"])/(["HA"]) = (2.96 × 10^"-3" × 2.96 × 10^"-3")/0.50 = 1.75 × 10^"-5"#

The reaction quotient #Q# is constant.