Methanoic Acid's dissociation constant

[josephcollins]

I need some help with these, anyone point me in the right direction please?

The pH of 0.04moldm-3 methanoic acid, HCOOH, is 2.59 at 25 celsius. Calculate the acid dissociation constant at this temperature

Given that the ionic product of water is 1.0 times 10^-14 mol2dm-6 at 298K, calculate to 3 significant figures the pH at this temperature of a 0.0500M solution of sodium hydroxide.

[chem_tr]

Hello,

0,04 M of formic acid has pH 2,59, as you gave us.

HCOOH (0,4-x) ---> H+ (x) + HCOO- (x)

We should carefully think of this problem first by considering the additional x in (0,4-x); and by omitting the x also. Note that we know x, 10-2,59. It means that the dissociation is about one hundredth; if it would be around 10%, then we should consider it. It won't be any problem if we omit the x in (0,4-x), I hope we've settled this.

Now the equilibrium constant gives us the dissociation constant (sorry for not even hearing Latex before :smile:):

K=(10-2,59*2)/0,4=1,652*10-5. The pKa will be found as 4,78.

About second question, you gave us 0,0500 M of NaOH and want pH as x.xxx. So, the contribution of water may be important. Again, if it is beyond 10%, it can be omitted.

NaOH (5,000*10-2) --> Na+ (5,000*10-2) + OH- (5,000*10-2) as it is fully ionized.

H2O --> H+ (1,000*10-7) + OH- (1,000*10-7), ionic product allows me to write these. As you see that they are well beyond 5,000*10-2, it can be omitted, since there is a five hundred thousand-fold difference.

The overall pH is (with enough precision) found to be 12,699. If you want to add water's hydroxide to it, you'd find the same, as 3 significant figures are not affected with this very small addition.

Regards, chem_tr