Question

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?

Answer 1

`(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`.