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
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
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 = ?
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 = ?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.