Help please!
How many moles of NaOH must be added to 1.0 L of 3.0 M HC2H3O2 to produce a solution buffered at each pH?
1) pH= 3.78
2) PH= 6.52
Acetic acid is HAc.
mols HAc = 1L x 3M = 3 mols.
..........HAc + OH^- ==> A^- + H2O
initial...3......0.......0.....0
add...............x...............
change.....-x.....-x......x
equil.......3-x....0.......x
Now substitute into Henderson-Hasselbalch equation and solve for for x
pH = pKa + log(base)/(acid)
To calculate the moles of NaOH required to produce a solution buffered at a specific pH, we need to consider the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([salt]/[acid])
Where pH is the desired pH, pKa is the acid dissociation constant, [salt] is the concentration of the salt, and [acid] is the concentration of the acid.
1) For pH = 3.78:
First, we need to find the pKa value for acetic acid (HC2H3O2). The pKa value of acetic acid is approximately 4.76.
pH = pKa + log([salt]/[acid])
3.78 = 4.76 + log([salt]/3.0)
Rearranging the equation:
log([salt]/3.0) = 3.78 - 4.76
log([salt]/3.0) = -0.98
To remove the logarithm, we take the antilog of both sides:
[salt]/3.0 = 10^(-0.98)
[salt]/3.0 = 0.108
Now, we can find the concentration of the salt ([salt]):
[salt] = (0.108)(3.0)
[salt] ≈ 0.324 M
To calculate the moles of NaOH needed, we need to use the balanced equation between NaOH and acetic acid (HC2H3O2). The balanced equation is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that we need equal moles of NaOH and HC2H3O2 to form the salt CH3COONa. Hence, the moles of NaOH required will be the same as the moles of HC2H3O2 in the solution, which is 3.0 moles per liter.
Therefore, 3.0 moles of NaOH should be added to 1.0 L of 3.0 M HC2H3O2 to produce a solution buffered at pH 3.78.
2) For pH = 6.52:
Using the same equation, we can set up the equation as:
pH = pKa + log([salt]/[acid])
6.52 = pKa + log([salt]/3)
Rearranging the equation:
log([salt]/3) = 6.52 - pKa
Now, we can find the concentration of the salt ([salt]) at pH 6.52. To do this, we need to know the pKa value of acetic acid. The pKa value of acetic acid is approximately 4.76.
log([salt]/3) = 6.52 - 4.76
log([salt]/3) = 1.76
To remove the logarithm, we take the antilog of both sides:
[salt]/3 = 10^(1.76)
[salt]/3 ≈ 58.04
Now, we can find the concentration of the salt ([salt]):
[salt] = (58.04)(3)
[salt] ≈ 174.12 M
To calculate the moles of NaOH needed, we use the balanced equation between NaOH and acetic acid (HC2H3O2):
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that we need equal moles of NaOH and HC2H3O2 to form the salt CH3COONa. Hence, the moles of NaOH required will be the same as the moles of HC2H3O2 in the solution, which is 3 moles per liter.
Therefore, 3 moles of NaOH should be added to 1.0 L of 3.0 M HC2H3O2 to produce a solution buffered at pH 6.52.