Asked by Ava

How many grams of sodium acetate, NaC2H3O2, would have to be added to 5.00 L of 0.65 M acetic acid (pKa 4.74) to make the solution a buffer for pH 4.00?

I did this but it's not the right answer:
4 =4.74 + log(A/.65)
0.74=log(A/0.65)
10^-0.74= A/0.65
0.182 = A/0.65
0.118= A
0.118mol/5.00L = 0.0236mol/L
0.0236mol/L*82g/mol= 1.94g
Answered by DrBob222
How much do you know about this. Post your work so far and tell us what your sticking point is.I can help you through it.
Answered by DrBob222
You substituted M - 0.65 for acid, therefore, the 0.118 you have calculated is molarity acetate ion.
M = mols/L
0.118 = mols/5.00
mols = 0.591 <b>This is your error.</b>
grams = mols x molar mass or
0.591 x molar mass = ?
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