zoozooboo
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for the following question i got up to 2.25x10^-3 mol of 25.45cm3 of vinegar, but i just don't understand how to get the 10cm3 of og vinegar.
explanation greatly appreciated


A student plans to determine the concentration, in moldm–3, of CH3COOH in a bottle of vinegar. The student will carry out a titration with aqueous barium hydroxide, Ba(OH)2(aq).
The student’s method is outlined below.
• Dilute 10.0cm3 of vinegar from the bottle with distilled water and make the solution up to 250.0cm3.
• Add the diluted vinegar to the burette.
• Titrate 25.0cm3 volumes of 0.0450moldm–3 Ba(OH)2 with the diluted vinegar.
The mean titre of the diluted vinegar is 25.45 cm3.
The reaction in the student’s titration is shown below.
2CH3COOH(aq) + Ba(OH)2(aq) → (CH3COO)2Ba(aq) + 2H2O(l)
(i) Calculate the concentration, in moldm–3, of CH3COOH in the original bottle of vinegar
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username4221054
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You know the moles of CH3COOH in the 25.45cm3 of vinegar.
You also know the volume (which is 25.45cm3).
Use the n=cv formula to calculate the concentration of CH3COOH.

The concentration of 25.45cm3 is the same as the concentration in the 250cm3 as the 25.45cm3 titre came from using the 250cm3 vinegar. So now you know the concentration and volume of the 250cm3 vinegar. Use n=cv to calculate the moles of the 250cm3 vinegar.

The moles of the 250cm3 vinegar is the same as the moles of the 10cm3 vinegar as diluting something doesn't change the moles of CH3COOH (you aren't adding or removing any CH3COOH, you're just adding water). So now you know the moles and volume of the 10cm3 vinegar. You use n=cv to calculate the concentration of the 10cm3 vinegar.

Hope this helped.
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zoozooboo
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(Original post by earthworm206)
You know the moles of CH3COOH in the 25.45cm3 of vinegar.
You also know the volume (which is 25.45cm3).
Use the n=cv formula to calculate the concentration of CH3COOH.

The concentration of 25.45cm3 is the same as the concentration in the 250cm3 as the 25.45cm3 titre came from using the 250cm3 vinegar. So now you know the concentration and volume of the 250cm3 vinegar. Use n=cv to calculate the moles of the 250cm3 vinegar.

The moles of the 250cm3 vinegar is the same as the moles of the 10cm3 vinegar as diluting something doesn't change the moles of CH3COOH (you aren't adding or removing any CH3COOH, you're just adding water). So now you know the moles and volume of the 10cm3 vinegar. You use n=cv to calculate the concentration of the 10cm3 vinegar.

Hope this helped.
Thanks alot!
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amani24
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(Original post by username4221054)
You know the moles of CH3COOH in the 25.45cm3 of vinegar.
You also know the volume (which is 25.45cm3).
Use the n=cv formula to calculate the concentration of CH3COOH.

The concentration of 25.45cm3 is the same as the concentration in the 250cm3 as the 25.45cm3 titre came from using the 250cm3 vinegar. So now you know the concentration and volume of the 250cm3 vinegar. Use n=cv to calculate the moles of the 250cm3 vinegar.

The moles of the 250cm3 vinegar is the same as the moles of the 10cm3 vinegar as diluting something doesn't change the moles of CH3COOH (you aren't adding or removing any CH3COOH, you're just adding water). So now you know the moles and volume of the 10cm3 vinegar. You use n=cv to calculate the concentration of the 10cm3 vinegar.

Hope this helped.
hi,
i was just doing this question and i couldn’t understand it for the life of me and your response really helped me.

however, i don’t really understand the last part of the calculation. i used n=cv to find the moles of 250cm3 vinegar. the concentration is 0.0884moldm-3. the mark scheme says to do 0.0884 x 250, but i don’t understand why i multiply by 250cm3 instead of 0.25dm3?
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deskochan
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(Original post by amani24)
hi,
i was just doing this question and i couldn’t understand it for the life of me and your response really helped me.

however, i don’t really understand the last part of the calculation. i used n=cv to find the moles of 250cm3 vinegar. the concentration is 0.0884moldm-3. the mark scheme says to do 0.0884 x 250, but i don’t understand why i multiply by 250cm3 instead of 0.25dm3?
We need to focus on the process and the number of moles:
1. 10cm3 vinegar is diluted to 250cm3 in the volumetric flask (they are the same number of moles in acid);
2. 25.45cm3 diluted vinegar is in the burette. and 1 & 2 are a different number of moles in acid and need to multiply 25.45/250 factor (9.823) from 2 to 1
Thus, the number of moles of acid in 25.45cm3 is 2.25x10^-3 mol and then needs to multiply 2.25x10^-3 x 9.823 in 250cm3
Finally concentration of the 10cm3 vinegar is (2.25x10^-3 x 9.823) x1000/10 moldm-3.
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lottitian
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(Original post by username4221054)
You know the moles of CH3COOH in the 25.45cm3 of vinegar.
You also know the volume (which is 25.45cm3).
Use the n=cv formula to calculate the concentration of CH3COOH.

The concentration of 25.45cm3 is the same as the concentration in the 250cm3 as the 25.45cm3 titre came from using the 250cm3 vinegar. So now you know the concentration and volume of the 250cm3 vinegar. Use n=cv to calculate the moles of the 250cm3 vinegar.

The moles of the 250cm3 vinegar is the same as the moles of the 10cm3 vinegar as diluting something doesn't change the moles of CH3COOH (you aren't adding or removing any CH3COOH, you're just adding water). So now you know the moles and volume of the 10cm3 vinegar. You use n=cv to calculate the concentration of the 10cm3 vinegar.

Hope this helped.
where are you getting the moles of CH3COOH from?
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