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I think you meant .2M, right?


PH=pKa+log(Base/Acid) pKa=-log(Ka)= 4.74.2=4.7+log(x/.2)

-.5=log(x/2)

10^-.5=x/.2

x=.0632M


.0632M X 1L X 82.03g/mol = 5.19g


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Q: How many grams of sodium acetate molar mass 82.03 must be added to 1.00 L pf a 200 M acetic acid solution to form a buffer of 4.20?
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