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Acids Question



Hey I've done this question and just want to make sure I followed it correctly. I can't get the markscheme as the OCR site is down and my files have been mixed up but here's my method:

n=m/M so do 72.9/45 = 1.58mol

apply n=c*v/1000 to get the concentration by rearranging therefore c = 2.10moldm^-3

Apply the Ka formula:
Ka = [CH3OO-][H+]/[CH3COOH]

Plugged the values in and got [H+] to be 5.97*10^-3
Converted it into pH and got the pH of 2.22

If anyone can be kind enough to check it for me, I'll really appreciate it :smile:
Original post by LifeIsGood


Hey I've done this question and just want to make sure I followed it correctly. I can't get the markscheme as the OCR site is down and my files have been mixed up but here's my method:

n=m/M so do 72.9/45 = 1.58mol

apply n=c*v/1000 to get the concentration by rearranging therefore c = 2.10moldm^-3

Apply the Ka formula:
Ka = [CH3OO-][H+]/[CH3COOH]

Plugged the values in and got [H+] to be 5.97*10^-3
Converted it into pH and got the pH of 2.22

If anyone can be kind enough to check it for me, I'll really appreciate it :smile:


Error in red.

NOTE : Your method is correct.... Just change that small error in the beginning :wink:
(edited 11 years ago)
Original post by LifeIsGood


Hey I've done this question and just want to make sure I followed it correctly. I can't get the markscheme as the OCR site is down and my files have been mixed up but here's my method:

n=m/M so do 72.9/45 = 1.58mol


the relative mass of ethanol is 46

79.2/46 = 1.72 mol



apply n=c*v/1000 to get the concentration by rearranging therefore c = 2.10moldm^-3


if 1.72 mol is in 0.75 litres the concentration is 1.72/.75 = 2.30 M




Apply the Ka formula:
Ka = [CH3OO-][H+]/[CH3COOH]

Plugged the values in and got [H+] to be 5.97*10^-3
Converted it into pH and got the pH of 2.22

If anyone can be kind enough to check it for me, I'll really appreciate it :smile:
Reply 3
Avatar for LifeIsGood
LifeIsGood
OP
12
Thanks for the quick replies! Small error on my part, glad I got it right. Thanks again!

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