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
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
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