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
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 = ?
To calculate the number of grams of sodium acetate needed to make the solution a buffer for pH 4.00, we need to use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the weak acid and its conjugate base. The equation is given by:
pH = pKa + log([A-]/[HA]),
where [A-] is the concentration of the conjugate base (sodium acetate in this case) and [HA] is the concentration of the weak acid (acetic acid).
Given that the pH is 4.00 and the pKa is 4.74, we can rearrange the equation to solve for [A-]:
4.00 = 4.74 + log([A-]/0.65),
-0.74 = log([A-]/0.65).
To obtain the value of [A-]/0.65, we can raise 10 to the power of -0.74:
[A-]/0.65 = 10^(-0.74),
[A-]/0.65 = 0.182.
To find the concentration of [A-], we multiply 0.182 by 0.65:
[A-] = 0.182 * 0.65,
[A-] = 0.118 M.
Now, to determine the number of moles of sodium acetate needed, we multiply the concentration by the volume of the solution:
0.118 mol/L * 5.00 L = 0.59 mol.
Finally, we can convert moles to grams by multiplying by the molar mass of sodium acetate (82 g/mol):
0.59 mol * 82 g/mol = 48.38 g.
Therefore, approximately 48.38 grams of sodium acetate would need to be added to 5.00 L of 0.65 M acetic acid to create a buffer solution with a pH of 4.00.