How many grams of sodium acetate were added ?

Question

To make a solution with a pH = 4.20 a student used the following procedure :
a certain amount of sodium acetate along with 0.290 moles of acetic acid is added to enough water to make a solution of 1.00 L
How many grams of sodium acetate were added ?

Ka = 1.80 x 10-5

Answer 1

pH = 4.20 >> Concentration H+ =0.0000631 M

Initial concentration CH3COOH = 0.290 M

Ka = 1.8 10^-5 = (0.0000631)( x + 0.0000631) / 0.290 -x

x = 0.0827M = concentration CH3COO-

In 1 liter mole CH3COO- = 0.0827 mole

MM CH3COONa = 80 g/mol

Grams of sodium acetate added = 0.0827 (80)= 6.62

Answer 2

Use the Henderson- Hasselbach equation

pH = pka + log (salt/acid)

4.20 = 4.74 + log (salt)/0.290)
-0.54 = log(salt/0.290)
10^-.54 = salt/0.290
.288 x .290 =salt
salt = 0.084 mol
grams = 0.084mol x 82g/mol = 6.90 g (sig fig)

Answer 3

you may quite attempt to remedy for conjugate base concentration utilising you Ka fee. set up this equation: Ka = [H3O][A]/[HA] -----> you have already got [HA], and additionally you gets [H3O] concentration by doing 10 raised to -3.9, which could supply you one million.26e-4. Now remedy for [A] utilising this equation, could be .0429M, convert that to mols (multiplying by Liters) could get .0429mols, multiply by molar mass (82g) and that would desire to offer you 3.52g. in basic terms examine the calculations incase I made a mistake.

Answer 4

pH = pka + log([A-]/[HA])
pka = -log(ka)
pH = log([A-]/[HA]) -log(ka)
so
4.20 = log([A-]/.290M) - log(1.80 * 10^-5)
-.544727495 = log([A-])/.290M)
10 ^ (-.544727495) = [A-]/.290M
[A-] = 0.082731425 M
since 1 L of solution is used, molarity = moles
0.082731425 mol * 82.03 g/mol
6.786458793 grams NaAcetate
rate my answer as best please