A 0.500M solution of a weak acid, HX, has a ka=3.26x10^-5 a) What is the pH of this solution? b) What is the percent ionization? c) What would be the pH be in a solution containing the strong electrolyte, 0.2M AlX3

A 0.500M solution of a weak acid, HX, has a ka=3.26x10^-5
a) What is the pH of this solution?
b) What is the percent ionization?
c) What would be the pH be in a solution containing the strong electrolyte, 0.2M AlX3?
d) What is the percent ionization in part C?
e) Does the presence of a strong electrolyte with a common ion increase or decrease the pH?

1 Answer
Jun 11, 2016

Answer:

#(a)" "2.4#

#(b)" " 0.8%#

#(c)" "4.56#

#(d)" "0.0054%#

#(e)" ""increase"#

Explanation:

#(a)#

To find the pH we can set up an #"ICE"# table:

#" "HX_((aq))" "rightleftharpoons" "H_((aq))^(+)" "+" "X_((aq))^-#

#color(red)("I")" "0.5" "0" "0#

#color(red)("C")" " -x" "+x" "+x#

#color(red)("E")" "(0.5-x)" "x" "x#

The expression for #K_a# is:

#K_a=([H_((aq))^+][X_((aq))^-])/([HX_((aq))])#

#:.K_a=x^2/(0.5-x)=3.26xx10^(-5)" ""mol/l"#

Because of the small value of #K_a# we can assume that #x# is much smaller than #0.5# so we can assume #(0.5-x)rArr0.5#

#:.K_a=x^2/0.5=3.26xx10^-5#

#:.x^2=3.26xx10^(-5)xx0.5=1.63xx10^(-5)#

#:.x=sqrt(1.63xx10^(-5))=4.037xx10^(-3)" ""mol/l"#

#:.[H_((aq))^+]=4.037xx10^(-3)" ""mol/l"#

#pH=-log[H_((aq))^+]=-log(4.037xx10^(-3))#

#color(red)(pH=2.4)#

#(b)#

From an initial moles of #0.5# only #4.037xx10^(-3)# are ions so the percent ionisation #=(4.037xx10^(-3))/(0.5)xx100=color(red)(0.8%)#

#(c)#

Now the solution contains #0.2"M"# of #AlX_3#. As it is a strong electrolyte we can say that this a #0.6"M"# solution of #X_((aq))^-#

Now the #"ICE"# table looks like this:

#" "HX_((aq))" "rightleftharpoons" "H_((aq))^(+)" "+" "X_((aq))^-#

#color(red)("I")" "0.5" "0" "0.6#

#color(red)("C")" " -x" "+x" "+x#

#color(red)("E")" "(0.5-x)" "x" "(0.6+x)#

#:.K_a=(x(0.6+x))/(0.5-x)#

Again, because #x# is much smaller than #0.5# and #0.6# this simplifies to:

#K_a=(0.6x)/0.5#

#:.K_a=1.2x=3.26xx10^(-5)#

#:.x=(3.26xx10^(-5))/1.2=2.716xx10^(-5) =[H_((aq))^+]#

#pH=-log[H_((aq))^+]=-log(2.716xx10^(-5))#

#color(red)(pH=4.56)#

#(d)#

Now the percentage ionisation becomes:

#(2.716xx10^(-5))/(0.5)xx100=color(red)(0.0054%)#

#(e)#

You can see that adding extra #X_((aq))^-# has raised the #pH#.

This is an example of "The Common Ion Effect" which is a consequence of Le Chatelier's Principle.

Adding a large amount of #X_((aq))^-# has caused the position of equilibria to shift to the left, reducing the amount of #H_((aq))^+# ions causing an increase in #pH#.