You need to prepare an acetate buffer of pH 5.77 from a 0.671 M acetic acid...

Question

Question

You need to prepare an acetate buffer of pH 5.77 f

Answer 1

Answer #1

no of moles of CH3COOH = MOlarity* volume in L

= 0.671*0.625 = 0.4194 moles

PH = PKa + log[CH3COOK]/[CH3COOH]

5.77 = 4.76 + log[CH3COOK]/0.4194

5.77-4.76 = log[CH3COOK]/0.4194

1.01 = log[CH3COOK]/0.4194

[CH3COOK]/0.4194 = 10^1.01

[CH3COOK]/0.4194 = 10.23

[CH3COOK] = 10.23*0.4194

[CH3COOK] = 4.29

no of moles of CH3COOK = 4.29moles

no of moles of CH3COOK = no of moles of KOH

no of moles of KOH = 4.29 moles

volume of KOH = no of moles /molarity

= 4.29/2.12 = 2.023L = 2023ml = 2.023*10³ml>>>> answer

Answer 2

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