Question

i think i am overlooking something... ive been working on this so long.

Goal is to identify an unknown monoprotic acid. A 0.026 M solution of magnesium hydroxide is used in a titration. To begin, 0.018 grams of the unknown acid are dissolved in 200mL of water prior to the titration. What is the molar mass of the unknown acid?

Answer 1

Monoprotic acid = acid which has the formula HA.
2 moles of the unknown acid would thus react with 1 mole of Mg(OH)2.
What you are missing is the volume of the unknown acid solution and the Mg(OH)2 used in the titration - without these values, we cannot determine the exact concentration of the acid used (in M) and hence the Mr of the acid.

Answer 2

@aurora25 : Are you able to obtain these values from a virtual lab or something?

Answer 3

I have a titration curve graph. But i suppose I don't know how to integrate the information into the problem.

Answer 4

Please upload a screenshot of the graph here.

Answer 5

I have a multiple choice set...
1. 7.5 g/mol
2. 15 g/mol
3. 30 g/mol
4. 25 g/mol

Answer 6

We need to see your graph before answering your question.

Answer 7

@aurora25 : Could you attach the picture or scan of the titration curve?

Answer 8

hopefully this works

Answer 9

OK, so the endpoint is reached when 23ml of Mg(OH)2 is added to the unknown acid. I am presuming that you are using 25ml of the unknown acid solution you made up previously for the titration?

Answer 10

That was one of my issues... the prompt doesn't state how much to use.

Answer 11

Ok, assume that we are using 25ml of acid in this titration. If that's the case, then we would have:

Number of moles of Mg(OH)2 used = 0.026 x 23/1000 = 0.000598 moles
Since 1 mole of Mg(OH)2 reacts with 2 moles of the acid, we would have:
0.000598 moles of Mg(OH)2 reacts with 0.001196 moles of the acid.
So, 25ml of the acid solution contains 0.001196 moles of the acid => 200ml of the acid solution contains 0.009568 moles of the acid.
Mr of the acid = 0.018/0.009568 = 2
This does not make sense, which means that we cannot be using 25ml of the acid solution, or else your stated concentration of Mg(OH)2 is incorrect.

Answer 12

@aurora25 : Could you please scan the question and upload it here?

Answer 13

Ok here it is...
Also, im logging out for a few minutes to switch computers... but ill be back in a few.

Answer 14

@aurora25: OK, it seems that you've typed the question as it has been provided to you. There isn't much information in this question, so we are just going to assume that the volume of the acid solution used for the titration is 200ml.

So, this means that our previous solution becomes modified as follows:
200ml of the acid solution contains 0.001196 moles of the acid.
Mr of the acid = 0.018/0.001196 = 15g

Hence the molar mass of the acid is 15g/mol, so the answer is b.