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
2 answers
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.
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 This is your error.
grams = mols x molar mass or
0.591 x molar mass = ?
M = mols/L
0.118 = mols/5.00
mols = 0.591 This is your error.
grams = mols x molar mass or
0.591 x molar mass = ?
1 answer