Question

(You may use the Henderson-Hasselbalch equation to perform the following calculations, but you do not have...

(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.

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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