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?

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.