Help with this chemistry question? (A2, acid-base equilibria)

I'm not sure how to tackle this question- the answer is apparently 12.63 and all my answers keep falling within ~1 of that so I feel like I'm on the right lines but keep getting it wrong ☹️ help please!!!

"Calculate the pH of the solution formed when 50cm3 of 0.5moldm^-3 Ba(OH)2 is added to 20cm3 of 0.1moldm^-3 HCOOH (Ka = 1.78 x 10^-4)"

Reply 1

TypicalNerd

Original post

by sr.06

I'm not sure how to tackle this question- the answer is apparently 12.63 and all my answers keep falling within ~1 of that so I feel like I'm on the right lines but keep getting it wrong ☹️ help please!!!
"Calculate the pH of the solution formed when 50cm3 of 0.5moldm^-3 Ba(OH)2 is added to 20cm3 of 0.1moldm^-3 HCOOH (Ka = 1.78 x 10^-4)"

Let’s begin by writing an equation for the overall reaction. Although methanoic acid contains two hydrogens, only one is actually acidic (it’s not common at A level to deal with C-H bonds that can ionise):

Ba(OH)2 + 2HCOOH —> Ba(HCOO)2 + 2H2O

So for every 2 mol of HCOOH present, 1 mole of Ba(OH)2 reacts.

We are told that 50 cm^3 of 0.50 mol dm^-3 Ba(OH)2 is used, which is 50/1000 x 0.5 = 0.025 mol and that 20 cm^3 of 0.1 mol dm^-3 HCOOH is used, which is 20/1000 x 0.1 = 0.002 mol.

Clearly the Ba(OH)2 is in excess (so there was no need to give you the Ka value) and from the ratio of HCOOH used up to Ba(OH)2 used up (2:1), we can deduce that 0.001 mol of Ba(OH)2 was used up. Hence 0.024 mol still remains.

If we look at the volumes of solutions used and assume there is no change in volume upon completion of the reaction (e.g the total volume is 50 + 20 = 70 cm^3, or 0.07 dm^3), then [Ba(OH)2] = 0.024/0.07 = 0.3428… mol dm^-3.

But notice that 1 mole of Ba(OH)2 contains 2 mol of OH^-, so it follows that [OH^-] = 2 x 0.3428… = 0.6857… mol dm^-3

Using the fact Kw = [H^+][OH^-] and Kw = 10^-14 mol^2 dm^-6, you should get [H^+] = 1.45… x 10^-14 mol dm^-3, which corresponds to a pH of about 13.84.

I suspect the expected answer of 12.63 is spurious and likely comes from an incorrect entry into a calculator or a typo in the question.

Reply 2

Pigster

Original post

by sr.06

I'm not sure how to tackle this question- the answer is apparently 12.63 and all my answers keep falling within ~1 of that so I feel like I'm on the right lines but keep getting it wrong ☹️ help please!!!
"Calculate the pH of the solution formed when 50cm3 of 0.5moldm^-3 Ba(OH)2 is added to 20cm3 of 0.1moldm^-3 HCOOH (Ka = 1.78 x 10^-4)"

There is one thing missing from the Q is T.

pH of bases are dependant on temperature since the autodissociation of water (H2O <-> H+ + OH-) is endothermic and hence Kw increases with T.

I can easily find that at 100oC Kw = 51.3 x 10-14 mol2 dm-6. Plugging that into TN's explanation produces a final pH of 12.13.

The 'correct' answer of 12.63 is therefore possible assuming you have a T of about 70oC, when Kw is about 16 x 10-14 mol2 dm-6.

Reply 3

TypicalNerd

Original post

by Pigster

There is one thing missing from the Q is T.
pH of bases are dependant on temperature since the autodissociation of water (H2O <-> H+ + OH-) is endothermic and hence Kw increases with T.
I can easily find that at 100oC Kw = 51.3 x 10-14 mol2 dm-6. Plugging that into TN's explanation produces a final pH of 12.13.
The 'correct' answer of 12.63 is therefore possible assuming you have a T of about 70oC, when Kw is about 16 x 10-14 mol2 dm-6.

Whilst temperature is a valid point, I do believe they gave you the Ka of methanoic acid - the value given being measured under standard conditions and therefore at 298 K, which suggests the solution is at 298 K even if it’s an irrelevant piece of data for the calculations themselves.

Reply 4

Nitrotoluene

I give the complete solution at this point, I am not doing any harm.
Okay, here's a step-by-step explanation of how to calculate the pH after the reaction of barium hydroxide and formic acid:
We are going to add 50 cm^3 of 0.5 mol/dm^3 Ba(OH)2 to 20 cm^3 of 0.1 mol/dm^3 HCOOH. There are the details:
To start with, we calculate the moles for each of the compounds:
Ba(OH)2: (0.5 mol/dm^3) x (0.050 dm^3) = 0.025 mol
HCOOH: (0.1 mol/dm^3) x (0.020 dm^3) = 0.002 mol
Now, Ba(OH)2 splits into Ba^2+ and 2 OH^- ions. Therefore, 0.025 mol of Ba(OH)2 gives us the following:
OH^- moles = 2 x 0.025 mol = 0.050 mol
Then, the OH^- reacts with the HCOOH (formic acid):
HCOOH + OH^- <==> HCOO- + H2O
When we have 0.002 mol of HCOOH and 0.050 mol of OH^-, completely , the HCOOH will react , which will result in:
HCOO^- moles = 0.002 mol
Remaining OH^- moles = 0.050 mol - 0.002 mol = 0.048 mol
The whole volume equals (50 cm^3 + 20 cm^3) = 70 cm^3 = 0.070 dm^3
Hence, the OH^- concentration will be:
[OH^-] = (0.048 mol) / (0.070 dm^3) = about 0.686 mol/dm^3
Now let's check the pOH:
pOH = -log[OH^-] = -log(0.686) = about 0.163
And the pH:
pH = 14 - pOH = 14 - 0.163 = about 13.837
Using these steps you will find that the pH of the solution is about 13.84.
Here is my 2 cents!

(edited 9 months ago)

Reply 5

sr.06