Calculate the correct volumes of the 0.1M acetic acid solution and the 0.1M sodium acetate solution needed to make 50mL of a buffer solution with a pH vaule of 5.00. (Ka = 1.8x10^-5)
You should use the Henderson-Hasselbalc equation.
To calculate the volumes of acetic acid and sodium acetate needed to make a buffer solution with a pH of 5.00, you can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
First, let's determine the ratio of [A-] / [HA] using the pH and pKa values:
pH = 5.00
pKa = -log10(Ka) = -log10(1.8x10^-5) = 4.75
pH = pKa + log ([A-] / [HA])
5.00 = 4.75 + log ([A-] / [HA])
0.25 = log ([A-] / [HA])
Using the logarithmic equation, we can convert the equation into exponent form:
10^0.25 = ([A-] / [HA])
Solving for the ratio of [A-] / [HA]:
[A-] / [HA] = 10^0.25 = 1.78
Now, let's calculate the volumes of acetic acid and sodium acetate needed.
Let's assume:
V1 = volume of 0.1M acetic acid solution (in mL)
V2 = volume of 0.1M sodium acetate solution (in mL)
Since you want a total volume of 50 mL, we can express the volume of sodium acetate in terms of the acetic acid volume:
V2 = 50 - V1
Using the molarities and volumes of the solutions, we can set up the following equation based on the definition of molarity:
Molarity = moles / volume (in liters)
For acetic acid:
0.1 M = moles of acetic acid / (V1/1000 L)
moles of acetic acid = 0.1 * (V1/1000)
For sodium acetate:
0.1 M = moles of sodium acetate / (V2/1000 L)
moles of sodium acetate = 0.1 * (V2/1000)
Since acetic acid is a weak acid and sodium acetate is a salt, we can assume that the number of moles of acetic acid is equal to the number of moles of acetate ions in the sodium acetate solution, based on their molar ratios. This allows us to set up an equation:
moles of acetic acid = moles of sodium acetate
0.1 * (V1/1000) = 0.1 * (V2/1000)
Simplifying the equation:
V2 = V1
Now we can calculate the volumes:
V2 = 50 - V1
V2 = V1
Substituting V2 = V1 into the previous equation:
0.1 * (V1/1000) = 0.1 * (V1/1000)
V1/1000 = V1/1000
V1 = V1
Therefore, the volumes of both the acetic acid and sodium acetate solutions are equal, and each should be 25mL.
To calculate the volumes of acetic acid and sodium acetate needed to make the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH of the buffer solution
pKa = -log(Ka), Ka being the acid dissociation constant
[A-] = concentration of the conjugate base (sodium acetate)
[HA] = concentration of the weak acid (acetic acid)
Now, let's plug in the given values:
pH = 5.00
pKa = -log(1.8x10^-5)
We need to rearrange the equation to solve for the ratio [A-]/[HA].
Taking the antilog of both sides, we get:
[A-]/[HA] = 10^(pH - pKa)
Substituting the values:
[A-]/[HA] = 10^(5.00 - (-log(1.8x10^-5)))
[A-]/[HA] = 10^(5.00 + 4.74)
[A-]/[HA] = 10^9.74
Now, we know that the sum of the volumes of acetic acid and sodium acetate solutions must be equal to 50 mL. Let's represent the volume of acetic acid as "Va" and the volume of sodium acetate as "Vs."
Va + Vs = 50 mL
Since we're dealing with molar solutions, we can use the formula:
n = C * V
Where:
n = number of moles
C = concentration (in mol/L)
V = volume (in L)
For acetic acid:
n(acetic acid) = C(acetic acid) * Va
For sodium acetate:
n(sodium acetate) = C(sodium acetate) * Vs
As we want the concentrations of both solutions to be 0.1 M, we can simplify for the sake of calculation:
n(acetic acid) = Va
n(sodium acetate) = Vs
Now, let's find the concentrations of acetic acid and sodium acetate based on the number of moles:
C(acetic acid) = n(acetic acid) / Va
C(sodium acetate) = n(sodium acetate) / Vs
From the Henderson-Hasselbalch equation, we know that [A-]/[HA] = 10^9.74. Since sodium acetate dissociates into sodium ions (Na+) and acetate ions (A-), we can conclude that the concentration of [A-] is equal to C(sodium acetate). Similarly, the concentration of [HA] is equal to C(acetic acid).
Therefore:
C(A-) = 10^9.74 * C(HA)
C(HA) = C(A-) / 10^9.74
Now, we have enough information to solve for the volumes Va and Vs.
Let's consider that initial concentration is the same as the final concentration after mixing:
C(initial acetic acid) * Va (initial acetic acid) + C(initial sodium acetate) * Vs (initial sodium acetate) = C(final) * (Va + Vs)
Since the initial concentrations are 0.1 M, we can rewrite the equation as follows:
(0.1 * Va) + (0.1 * Vs) = 0.1 * 50 mL
Simplifying:
0.1Va + 0.1Vs = 5 mL
Rearranging the equation:
Va + Vs = 50 mL
From the earlier equation, we have C(HA) = C(A-) / 10^9.74, so we can substitute those values into the equation:
(0.1 * Va) + (0.1 * 10^9.74 * Va) = 5 mL
Simplifying:
0.1Va + (10^8.374 * Va) = 5 mL
Combining like terms:
1.1Va = 5 mL
Solving for Va:
Va = 5 mL / 1.1
Va = 4.55 mL (approximately)
To find Vs, we can substitute the value of Va into the equation Va + Vs = 50 mL:
4.55 mL + Vs = 50 mL
Solving for Vs:
Vs = 50 mL - 4.55 mL
Vs = 45.45 mL (approximately)
Therefore, to make a 50 mL buffer solution with a pH of 5.00 using 0.1 M acetic acid and 0.1 M sodium acetate solutions, you would need approximately 4.55 mL of acetic acid and 45.45 mL of sodium acetate solution.