Question

2.5x10-3 moles of solid Mg(OH)2 is added to a 100.0 mL sample of 0.250 M aqueous chlorous acid solution. Ignore the volume change associated with the added solid. Find the pH of the solution?

Use the information below to answer the following question:
2.5x10-3 moles of solid Mg(OH)2 is added to a 100.0 mL sample of 0.250 M
aqueous chlorous acid solution. Ignore the volume change associated with the
added solid. Find the pH of the solution.

Answer 1

`sf(pH=1.6)`

Explanation:

The base is neutralised by the acid:

`sf(Mg(OH)_2+2HClO_2rarrMg(ClO_2)_2+2H_2O)`

The number of moles of chlorous acid is given by:

`sf(n_(HClO_2)=cxxv=0.250xx100.0/1000=0.0250)`

We are told that:

`sf(n_(Mg(OH)_2)=0.0025)`

From the stoichiometry of the reaction we can see that the number of moles of `sf(HClO_2)` consumed must be `sf(0.0025 xx 2 = 0.005)`.

So the number remaining is given by:

`sf(n_(HClO_2)=0.025-0.005=0.02)`

We can also say that the number of moles of `sf(ClO_2^-)` formed = `sf(0.0025 xx 2 = 0.005)`

We have now created an acid buffer since we have a weak acid in combination with its co - base.

Chlorous acid dissociates:

`sf(HClO_2rightleftharpoonsH^(+)+ClO_2^-)`

For which:

`sf(K_a=([H^+][ClO_2^-])/([HClO_2])=10^(-12)color(white)(x)"mol/l")`

The concentrations are those at equilibrium.

To find `sf([H^+])` and hence the pH we need to set up an ICE table based on concentration.

Since we are told the volume is 100 ml = 0.1 L then the initial concentrations become:

`sf([HClO_2]=0.02/0.1=0.2color(white)(x)"mol/l"`

`sf([HClO_2^-]=0.005/0.1=0.05color(white)(x)"mol/l")`

`sf([H^+]=0color(white)(x)"mol/l")`

`""` `" "sf(HClO_2" "rightleftharpoons" "H^(+)" "+" "ClO_2^-)`

`sf(color(red)(I)" "0.2" "0" "0.05)`

`sf(color(red)(C)" "-x" "+x" "+x)`

`sf(color(red)(E)" "(0.2-x)" "x" "(0.05+x))`

`:.` `sf(K_a=(x(0.05+x))/((0.2-x))=10^(-2)`

At this point it is common to assume that, because the dissociation is so small then `sf((0.05+x)rArr0.05)` and `sf((0.2-x)rArr0.2)`.

However, we can't do this in this case as the size of `sf(K_a)` means that `sf(x)` is significant.

If we multiply out the expression we get:

`sf(x^2+0.06x-0.002=0)`

This is a quadratic equation which can be solved for `sf(x)` using the quadratic formula. I won't go into this here but this gives, ignoring the absurd root:

`sf(x=0.02385color(white)(x)"mol/l"=[H^+])`

`sf(pH=-log[H^+]=-log(0.02385)=1.6)`

If we make the assumption that `sf((0.05+x)rArr0.05)` and `sf((0.2-x)rArr0.2)` then `sf(x(0.05)/0.2=0.01)` which gives `sf(x=0.4)` for which `sf(pH=1.4)` which is significantly lower.