[DesertRose]

Hi, I attempted this question, but i'm not sure if i'm going about it right:

A chemist extracts the methanoic acid from 40 ants and adds water to make 25.oo cm³⁺ of a solution Y. He then titrates this solution with a 0.050 mol /dm3 standard solution of sodium hydroxide, NaOH(aq), a strong alkali, to determine the concentration of HCOOH in Solution Y.

1) Using Bronsted-Lowry theory, explain what is meant by a strong base.
2) Calculate the pH of the standard NaOH solution. 
3) The concentration of methanoic acid in Solution  Y is found to be 6.0 x 10^3-.    Calculate the pH of Solution Y

For 1) I know that a strong base is one that completely ionises in aqueous solution, and the Bronsted-Lowry theory states that a base is a proton acceptor, but how can a strong base be explained using the Bronsted-Lowry theory?
Also, just making sure, when the term 'aqueous solution' or just 'aqueous' is used, does that always refer to water?

2) i found the pH to be 8.2, however i am not sure whether this answer is correct and if the  method i used is correct either. This is my method: I found Kb using the formula (Ka x Kb = 1.0 x 10^-14). i then substituted this value and the [NaOH] into the Kb expression. I solved for [OH-]2. Then found the value for pOH. Using the formula pH + pOH = 14, i got the pH to be 8.2. Is this correct?

3) I wrote the equation: HCOOH(aq) + H₂O(l) H₃O⁺(aq) + HCOO^-(aq)
From the ratio, [H₃O⁺] = [HCOOH] = 6.0 x 10^3-
Using this i found the pH to be 2.2

Is this correct?
Any assistance will be greatly appreciated. Thank you!

[Borek]

For 1) I know that a strong base is one that completely ionises in aqueous solution, and the Bronsted-Lowry theory states that a base is a proton acceptor, but how can a strong base be explained using the Bronsted-Lowry theory?

Agreed, question is rather strange.

Also, just making sure, when the term 'aqueous solution' or just 'aqueous' is used, does that always refer to water?

Yes. 'Aqua' means 'water' in Latin.

2) i found the pH to be 8.2, however i am not sure whether this answer is correct and if the  method i used is correct either. This is my method: I found Kb using the formula (Ka x Kb = 1.0 x 10^-14). i then substituted this value and the [NaOH] into the Kb expression. I solved for [OH-]2. Then found the value for pOH. Using the formula pH + pOH = 14, i got the pH to be 8.2. Is this correct?

No. If NaOH is fully dissociated, you don't need to use Kb. COncentration of OH^- is that of the NaOH, pretty simple. Besides, I have no idea how you calculated it from the equation you gave, if you are not given Ka.

3) I wrote the equation: HCOOH(aq) + H₂O(l) H₃O⁺(aq) + HCOO^-(aq)
From the ratio, [H₃O⁺] = [HCOOH] = 6.0 x 10^3-

During titration you have determined total concentration of formic acid (HCOOH and HCOO^-) to be equal 6×10^-3. This is not concentration of HCOOH. Besides, I don't understand where you got  [H₃O⁺] = [HCOOH] from.

Using this i found the pH to be 2.2

Is this correct?

I am afraid no. You should use determined concentration and formic acid Ka (from tables) for calculations.

[DesertRose]

Thank you for your feedback Borek.
I did the question over and for part 2) i used the formula [OH-][H+]= 1.0 x 10^-14 . Substituting [OH-] as 0.050 mol/dm3, i found for [H+] then found for the pH. I got 12.7 as my answer?
So the only time you can say that the concentration of each of the ion products in a dissociation reaction is equal to the reactant substance, is when the reactant substance completely ionises only (i.e. only for strong acids and bases?)

 In part 3) i thought that the conc. of the products = concentration of reactants ( i did not know that it was different for strong and weak solutions) Hence i did this:

3) I wrote the equation: HCOOH(aq) + H₂O(l) H₃O⁺(aq) + HCOO^-(aq)
From the ratio, [H₃O⁺] = [HCOOH] = 6.0 x 10^3-

During titration you have determined total concentration of formic acid (HCOOH and HCOO^-) to be equal 6×10^-3. This is not concentration of HCOOH. Besides, I don't understand where you got  [H₃O⁺] = [HCOOH] from.

So for a weak substance, when given the conc. of the reactant, you have to solve for the conc of the ions using Ka (for acid) or Kb (for base) and the conc. of the substance?

I did 3) over:
I was given Ka of methanoic acid in a an earlier part of the question as 1.6 x 10^-4.
Using Ka = [H3O+][HCOO-]/[HCOOH] i substituted [HCOOH] as 6.0 x 10^-3 and Ka as 1.6 x 10^-4. i solved for [H3O]squared, then for pH and got pH to be 3.0. Is this correct?

Just making sure, [HCOOH] was given as 6.0 x 10^-3. But how do you know if that is the value at equilibrium or at the start?

[Borek]

Substituting [OH-] as 0.050 mol/dm3, i found for [H+] then found for the pH. I got 12.7 as my answer?

OK

So the only time you can say that the concentration of each of the ion products in a dissociation reaction is equal to the reactant substance, is when the reactant substance completely ionises only (i.e. only for strong acids and bases?)

Yes.

So for a weak substance, when given the conc. of the reactant, you have to solve for the conc of the ions using Ka (for acid) or Kb (for base) and the conc. of the substance?

Sorry, I have problems following what you write, as it is not clear to me what "substance" and "reactant" refer to.

I was given Ka of methanoic acid in a an earlier part of the question as 1.6 x 10^-4.
Using Ka = [H3O+][HCOO-]/[HCOOH] i substituted [HCOOH] as 6.0 x 10^-3 and Ka as 1.6 x 10^-4. i solved for [H3O]squared, then for pH and got pH to be 3.0. Is this correct?

Just making sure, [HCOOH] was given as 6.0 x 10^-3. But how do you know if that is the value at equilibrium or at the start?

No, that's not the correct approach. As explained earlier, concentration determined by titration is an ANALYTICAL concentration, not EQUILIBRIUM concentration. So you have to solve problem that states "what is pH of 6×10^-3 M solution of formic acid", not "what is pH of a formic acid solution in which concentration of undissociated HCOOH is 6×10^-3 M".

[DesertRose]

I still do not understand what i did wrong for part 3) 

[Borek]

Formic acid dissociates, and solution contains HCOOH, HCOO^- and H⁺. When you titrate the solution you neutralize both H⁺ and HCOOH, so the concentration that you determined (6×10^-3 M) is not that of equilibrium HCOOH, but 6×10^-3 = [HCOOH] + [HCOO^-]. This in turn means you can't plug 6×10^-3 directly into Ka as if it was [HCOOH].

[DesertRose]

Ok, so here's what i did for 3)
i made an ICE table (so you always use the ICE table for weak acid or base?--and never for strong acid or base?) then substituted the value of Ka as 1.6 x 10^-4, and put it equal to
x squared / (6.0 x 10^-3)-x. since it is a weak acid, the value of (6.0 x 10^-3)-x is approx. equal to
(6.0 x 10^-3). i then solved for x squared then x , and calculate the pH as 3.0. Is this correct?

Just making sure, if this was a strong acid calculation instead, you can say that
6.0 x 10^-3 = [solution Y] = [H3O+] = [HCOO-] since it completely dissociates, but if, just say, instead of 1 mol H3O+ in the equation, we had 2 moles H3O+ then you say 2(6.0 x 10^-3) to calculate the concentration of H3O+ in the equation, and in the expression for Ka, that value will be squared...is that right? But in a weak acid, since it doesn't completely dissociates, you have to use an ICE table, and the concentration that they give you is actually the initial concentration?

[Borek]

i made an ICE table (so you always use the ICE table for weak acid or base?--and never for strong acid or base?)

You don't need ICE table for a strong acid/base.

then substituted the value of Ka as 1.6 x 10^-4, and put it equal to
x squared / (6.0 x 10^-3)-x. since it is a weak acid, the value of (6.0 x 10^-3)-x is approx. equal to
(6.0 x 10^-3). i then solved for x squared then x , and calculate the pH as 3.0. Is this correct?

Close to correct. After calculations you should check how precise was the assumption that 6.0×10^-3-x is approx. equal to 6.0×10^-3. There is a so called 5% rule of a thumb that says if one of the elements in the sum is smaller than 5% of the other, it can be ignored. Is x here smaller than 5% of 6.0×10^-3?

Just making sure, if this was a strong acid calculation instead, you can say that
6.0 x 10^-3 = [solution Y] = [H3O+] = [HCOO-] since it completely dissociates, but if, just say, instead of 1 mol H3O+ in the equation, we had 2 moles H3O+ then you say 2(6.0 x 10^-3) to calculate the concentration of H3O+ in the equation, and in the expression for Ka, that value will be squared...is that right?

Sorry, no idea what you are talking about. For a strong acid you assume it is 100% dissociated, period. No idea where does 1M H⁺ comes from in the solution that is 6.0×10^-3 M in acid.

[DesertRose]

No....x is not smaller than 5% of 6.0×10^-3....could i still leave my answer or because of this, the question has to be redone using the quadratic formula?

[Borek]

It would be better to redo the question using quadratics.

[DesertRose]

Ok, thank you very much!