what mass of sodium acetate should be dissolved in 1.00L of acetic acid, 0.150M concentration, in order to produce a buffer with pH = 4.5?. Ka of CH3COOH = 1.8x10^-5
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
Plug in pH needed, pKa for acetic acid, (HAc) for acid in the log term, and solve for base (acetate). That will be the concn of sodium acetate. For 1 L in the problem, that = moles and mols = grams/molar mass. Solve for grams.
To calculate the mass of sodium acetate needed to produce a buffer with pH 4.5, we need to use the Henderson-Hasselbalch equation. From the equation, we know the following:
pH = pKa + log ([A-]/[HA])
Where:
pH = desired pH of the buffer (4.5 in this case)
pKa = -log Ka (Ka is given as 1.8x10^-5, so pKa = -log (1.8x10^-5))
[A-] = concentration of the conjugate base (CH3COO-) in the buffer
[HA] = concentration of the weak acid (CH3COOH) in the buffer
First, let's calculate pKa:
pKa = -log (1.8x10^-5) = 4.74
Next, we'll determine the concentration ratio ([A-]/[HA]):
pH = pKa + log ([A-]/[HA])
4.5 = 4.74 + log ([A-]/[HA])
log ([A-]/[HA]) = 4.5 - 4.74
log ([A-]/[HA]) = -0.24
Now, we'll calculate ([A-]/[HA]) by taking the antilog of both sides:
[A-]/[HA] = antilog(-0.24)
[A-]/[HA] = 10^(-0.24)
Now, we can calculate the concentration of the acetate ion ([A-]) in the buffer by multiplying [A-]/[HA] by the concentration of the weak acid (0.150 M):
[A-] = (10^(-0.24)) * (0.150 M)
Next, we'll calculate the molar mass of sodium acetate (CH3COONa) using the atomic masses:
M(CH3COONa) = M(C) + M(H) + 3 * M(O) + M(Na)
M(C) = 12.01 g/mol
M(H) = 1.008 g/mol
M(O) = 16.00 g/mol
M(Na) = 22.99 g/mol
M(CH3COONa) = 12.01 + 1.008 + 3 * 16.00 + 22.99
Finally, we can calculate the mass of sodium acetate (CH3COONa) needed using the molar mass and the concentration of the acetate ion ([A-]):
Mass of CH3COONa = ([A-]) * Volume * Molar mass
Mass of CH3COONa = ([A-]) * 1.00 L * M(CH3COONa)
To determine the mass of sodium acetate needed to produce a buffer with a specific pH, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of the acid and its conjugate base. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH (4.5 in this case), pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid (1.8x10^-5 in this case), [A-] is the concentration of the conjugate base (acetate ion), and [HA] is the concentration of the weak acid (acetic acid).
We know the concentration of the acetic acid, which is 0.150 M. We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values:
[A-]/[HA] = 10^(4.5 - (-5))
[A-]/[HA] = 10^9.5
Now, since the concentration of the sodium acetate will be equal to the concentration of the acetate ion, we can use the relationship between concentration, volume, and moles to calculate the amount of sodium acetate needed.
The equation for concentration is:
C = n/V
Where C is the concentration, n is the number of moles, and V is the volume.
We can rearrange this equation to solve for n:
n = C * V
We know the volume V is 1.00 L. So, we need to calculate the concentration of sodium acetate required. From the previous step, we determined the ratio [A-]/[HA] to be 10^9.5. Since [A-] = [NaCH3COO], we can set up the equation as follows:
0.150 M (acetic acid) = [NaCH3COO]
Now, substitute the concentration and volume into the equation to calculate the number of moles:
n = (0.150 M) * (1.00 L)
n = 0.150 moles
Finally, we convert the number of moles to mass using the molar mass of sodium acetate:
Molar mass of sodium acetate = (22.99 g/mol for Na) + (12.01 g/mol for C) + (3 * 1.01 g/mol for H) + (2 * 16.00 g/mol for O) = 82.03 g/mol
Mass = n * molar mass
Mass = 0.150 moles * 82.03 g/mol
Mass = 12.3045 g
Therefore, you would need to dissolve approximately 12.3045 g of sodium acetate in 1.00 L of acetic acid solution to produce a buffer with a pH of 4.5.