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

(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.