Question
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
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
Answers
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.
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 = ?
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 = ?