Kc calculation from initial rates.

Original post by Pigster

You haven't mentioned 6-5

what do you mean by 6-5?

this is the thing I get stuck on I don't know how to change it concentration.

yes I know what kc equation is.

Reply 3

6-5 is the final result for the ch3ooh so then you can substitute your values into the kc equation.

Reply 4

Original post by Menakshelatte

what do you mean by 6-5?

this is the thing I get stuck on I don't know how to change it concentration.

yes I know what kc equation is.

You have the total volume is 1 dm^3 and you have values of mol for things, you should be able to calculate the concentration of each substance at equilibrium.

Reply 5

Remember that concentration equals the number of moles divided by the volume. In this case it is 1dm3.

Reply 6

Original post by Maannashon

Remember that concentration equals the number of moles divided by the volume. In this case it is 1dm3.

Original post by NDVA

You have the total volume is 1 dm^3

and you have values of mol for things, you should be able to calculate the concentration of each substance at equilibrium.

Original post by Maannashon

6-5 is the final result for the ch3ooh so then you can substitute your values into the kc equation.

ok then thanks to all of you guys.

in june 2013 paper volume of HCOOH = 200 cm and conc =3.2 mol dm.

so i did 200x3.2 / 100 to get moles. I got 0.64
but mark scheme says 0.24.

do you know why this is?

This is ph calculation btw.

Reply 7

Does the question give you the kc value of hcooh?

Reply 8

Original post by Maannashon

Does the question give you the kc value of hcooh?

no it gives you ka value.

1.70 × 10–4

Reply 9

Ah sorry thats what I meant. But it asks you to calculate the ph of the 200cm3 of 3.2 mol dm3 hcooh right?

Reply 10

Original post by Menakshelatte

Sorry, took me a while to look it up because my exam board is not OCR.

The first reaction is HCOOH + NaOH \rightarrow HCOONa + H_20, from this you can calculate how many moles of HCOOH reacted. If you took away this value from initial moles, which is 0.64, you would get 0.24, which is moles in equilibrium of the buffer solution.

(edited 8 years ago)

Reply 11

Original post by NDVA

Sorry, took me a while to look it up because my exam board is not OCR.

The first reaction is HCOOH + NaOH \rightarrow HCOONa + H_20

, from this you can calculate how many moles of HCOOH reacted. Then you take away this value from the initial moles, which is 0.64, you would get 0.24.

I did not undertand this part.

because they gave us 2 pieces of information in that question
1. 200 cm and 3.20 mol (concentration) hcooh
2.800 cm and 0.5 mol(concentration) Naoh.

when I calculated the moles for the second one, I got 0.4 directly. I did not have to do any of this taking away stuff.

and what is 0.24 the moles of exactly?

I dunno it's a bit confusing.
I am not doing this taking away thing for the second one.

hcooh i got 0.64

but then it's 0.64-0.4=0.24

I don't see where this 0.4 is coming from or why I'm doing this calculation.

Reply 12

Original post by Menakshelatte

I did not undertand this part.

because they gave us 2 pieces of information in that question
1. 200 cm and 3.20 mol (concentration) hcooh
2.800 cm and 0.5 mol(concentration) Naoh.

when I calculated the moles for the second one, I got 0.4 directly. I did not have to do any of this taking away stuff.

and what is 0.24 the moles of exactly?

I dunno it's a bit confusing.
I am not doing this taking away thing for the second one.

hcooh i got 0.64

but then it's 0.64-0.4=0.24

I don't see where this 0.4 is coming from or why I'm doing this calculation.

The initial moles of HCOOH is 0.64.

After the reaction HCOOH+NaOH \rightarrow HCOO^-Na^++H_20, moles of HCOOH that reacted with NaOH will be 0.4.

Hence moles of HCOOH remained = 0.64 - 0.4 = 0.24 mol :smile:

The equation of the buffer solution is then HCOOH \Leftrightarrow HCOO^- + H^+, with moles of HCOOH in this solution is 'moles remained' 0.24 mol.

Hope this is less confusing than my previous post!

(edited 8 years ago)

Reply 13

Original post by NDVA

The initial moles of HCOOH is 0.64.

After the reaction HCOOH+NaOH \rightarrow HCOONa+H_20

, mol of HCOOH that reacted with NaOH will be 0.4.

Hence mol of HCOOH remained = 0.64 - 0.4 = 0.24 mol :smile:

Hope this is less confusing than my previous post!

loooool I have made a note of all of the things you have said in my scrapbook : D

you have enlightened me so much Thank you : D

ahhh ok so 0.24 is the moles that have remained and we use that in out Ph calculation.

Reply 14

Original post by Menakshelatte

loooool I have made a note of all of the things you have said in my scrapbook : D

you have enlightened me so much Thank you : D

ahhh ok so 0.24 is the moles that have remained and we use that in out Ph calculation.

Yeah so basically one way to prepare a buffer solution you mix an acid of high concentration with an alkaline.

RCOOH + NaOH \rightarrow RCOO^-Na^+ + H_20

After this reaction, the remained high conc. acid will be in equilibrium with its salt (product of the above reaction), hence the equation:

RCOOH \Leftrightarrow RCOO^- + H^+

From this, K_a = \frac{conc. of salt \times conc. of H^+}{conc. of RCOOH} .

Hope this helps :smile:

(edited 8 years ago)

Reply 15