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Acids Question

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    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
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    (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
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    (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
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    Thanks for the quick replies! Small error on my part, glad I got it right. Thanks again!

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Updated: April 22, 2012
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