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# Acids Question

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1. 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
2. (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
Error in red.

NOTE : Your method is correct.... Just change that small error in the beginning
3. (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
4. Thanks for the quick replies! Small error on my part, glad I got it right. Thanks again!

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