OHNOGEM
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Hi can anyone help me solve buffer solution question, I don't understand how to answer them and why they are different to acid-base q's

Ka for weak acid HY is1.35x10^-5 at 25 degrees

A buffer solution is prepared by dissolving 0.0236 mol of the sale NaY in 50.0cm3of a 0.428 moldm-3 solution of the wean acid HY
Calculate PH
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sophieshoesmith
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I'm GCSE student but I'm assuming you mean salt not sale lol
Conc of 0.0236 mol in 50cm3 gives x20, a 0.472 molar solution for the salt

I googled how to do bufferpH and it said use the Henderson equation,

  • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]
    and using that I get
    1.35x10^-5 =[H+]*[0.472]/[0.428]

    so [H+]=1.35x10^-5/1.1 = 1.2*10^-5

    you can get the pH from that number by taking the log of it

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OHNOGEM
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Report Thread starter 7 years ago
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(Original post by sophieshoesmith)
I'm GCSE student but I'm assuming you mean salt not sale lol
Conc of 0.0236 mol in 50cm3 gives x20, a 0.472 molar solution for the salt

I googled how to do bufferpH and it said use the Henderson equation,

  • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]
    and using that I get
    1.35x10^-5 =[H+]*[0.472]/[0.428]

    so [H+]=1.35x10^-5/1.1 = 1.2*10^-5


    you can get the pH from that number by taking the log of it

Hey thanks for taking the time to do this; feel pretty embarrassed a GCSE student is helping me with my A2 work.. but hey-ho thanks again!
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