(You may use the Henderson-Hasselbalch equation to perform the following calcula

Question

(You may use the Henderson-Hasselbalch equation to perform the following calculations, but you do not have to.) The Ka of acetic acid is 1.8 10–5. Review your calculations with your TA or instructor before preparing the buffer solutions in lab. (Note: solid sodium acetate comes as a hydrate (NaC2H3O2·3H2O), and thus has a molar mass of 135.08 g/mol.) Buffer A: Calculate the mass of sodium acetate that must be added to make 100.0 mL of an acetic acid/acetate buffer at pH 4.00, given that you will use 5.00 mL of 1.0 M acetic acid.

Explanation / Answer

Volume of buffer A to be made=100 ml=100/1000 L=0.1L

pH=4.0

moles of acetic acid to be used=volume * molarity=5.0 ml*1.0M=5/1000 L* 1.0 mol/L=0.005 moles

using Henderson-hasselbach equation,

pH=pKa+log [conjugate base]/[acid]

pH=pka + log [sodium acetate]/[acetic acid]

ka=1.8*10^-5

pka=-log ka=-log (1.8*10^-5)=- (-5+0.255)=-(-4.745)=4.745

4=4.745+ log (mb/V)/(ma/V)

mb=moles of conjugate base

V=total volume of solution

ma=moles of acid

[sod acetate]=ma/V

[acetic acid]=0.005mol/V

4=4.745+ log (mb/0.005)

4-4.745= log (mb/0.005)

-0.745= log (mb/0.005)

10^(-0.745)=mb/0.005

0.180= mb/0.005

Mb=0.180*0.005=0.0009 moles

So mass of sod acetate=0.0009 moles*135.08 g/mol=0.1216g (answer)

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