Calculate the formal concentrations of acetic acid and sodium acetate necessary to prepare a buffer solution of pH 5 that is 0.20M in total acetate?
Why did the acetic acid go to school? To get a little buffer education!
Now, let's solve this problem step by step. The Henderson-Hasselbalch equation for a buffer solution is:
pH = pKa + log ([A-]/[HA])
Given that we want a pH of 5, we can rearrange the equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
The pKa value for acetic acid is approximately 4.76. Plugging in these values:
[A-]/[HA] = 10^(5 - 4.76) = 1.74
Now, let's assign variables to the concentrations of acetic acid and sodium acetate. Let [HA] be the concentration of acetic acid and [A-] be the concentration of sodium acetate.
Given that the total acetate concentration is 0.20 M, we have:
[HA] + [A-] = 0.20 M
Using the ratio we found earlier ([A-]/[HA] = 1.74), we can rewrite this equation as:
[HA] + 1.74[HA] = 0.20 M
Combining like terms:
2.74[HA] = 0.20 M
Dividing both sides by 2.74:
[HA] = 0.20 M / 2.74 = 0.073 M
And:
[A-] = 1.74 * [HA] = 1.74 * 0.073 M = 0.127 M
So the formal concentrations for acetic acid and sodium acetate in the buffer solution are approximately 0.073 M and 0.127 M, respectively.
To calculate the formal concentrations of acetic acid and sodium acetate necessary to prepare a buffer solution, we first need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA]),
where pH is the desired pH of the buffer solution, pKa is the dissociation constant of acetic acid (4.76), [A-] is the concentration of the conjugate base (in this case, sodium acetate), and [HA] is the concentration of the acid (in this case, acetic acid).
Given that the total concentration of acetate in the buffer solution is 0.20 M, we can express concentrations of acetic acid and sodium acetate in terms of x:
[HA] = x,
[A-] = 0.20 - x.
Substituting these values into the Henderson-Hasselbalch equation:
5 = 4.76 + log ((0.20 - x) / x).
Now, let's solve for x:
5 - 4.76 = log ((0.20 - x) / x).
0.24 = log ((0.20 - x) / x).
10^(0.24) = (0.20 - x) / x.
1.5849 = (0.20 - x) / x.
1.5849x = 0.20 - x.
1.5849x + x = 0.20.
2.5849x = 0.20.
x = 0.20 / 2.5849.
x = 0.0774 M.
Therefore, the formal concentration of acetic acid is 0.0774 M, and the formal concentration of sodium acetate is (0.20 - 0.0774) M = 0.1226 M.
To calculate the formal concentrations of acetic acid and sodium acetate necessary to prepare a buffer solution of pH 5, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the conjugate acid and conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base (acetate in this case), and [HA] is the concentration of the weak acid (acetic acid).
In this case, we want the buffer solution to have a pH of 5. Since acetic acid is a weak acid with a pKa of approximately 4.76, we can substitute the values into the Henderson-Hasselbalch equation:
5 = 4.76 + log([A-]/[HA])
Now, we need to solve for [A-]/[HA], which will give us the ratio of the concentrations of acetate to acetic acid. Rearranging the equation:
log([A-]/[HA]) = 5 - 4.76
log([A-]/[HA]) = 0.24
To remove the logarithm, we can take the antilog of both sides:
10^(log([A-]/[HA])) = 10^0.24
[A-]/[HA] = 1.58
Now, we know that the total acetate concentration ([A-] + [HA]) is 0.20 M, so we can express [A-] in terms of [HA]:
[A-] = 1.58[HA]
Substituting this into the equation for total acetate:
0.20 = 1.58[HA] + [HA]
0.20 = 2.58[HA]
Now, we can solve for [HA]:
[HA] = 0.20 / 2.58
[HA] ≈ 0.078 M
Finally, we can calculate [A-]:
[A-] = 1.58 * 0.078
[A-] ≈ 0.123 M
Therefore, to prepare a buffer solution of pH 5 that is 0.20 M in total acetate, you will need acetic acid with a concentration of approximately 0.078 M and sodium acetate with a concentration of approximately 0.123 M.
You need two equations to solve simultaneously. Equation 1 is the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
5 = pKa + log (base/acid)
Substitute and solve for ratio (base)/(acid).
Equation 2 is
(acid) + (base) = 0.2 M
Solve the two equations for (base) and (acid) and you have your answers. acetic acid is the acid and sodium acetate is the base.