If you need to prepare 250.0 mL of a pH 5.00 buffer tha
t has a total buffer concentration of acetic acid +
sodium acetate of 0.050 M, how many moles of each will you need to prepare the solution? Given solutions
of acetic acid and sodium acetate with concentrations of 0.10 M and pKa= 4.76, describe how to pre
pare this
buffer.
That's math, not chemistry. :).
To make it easy typing let me call the acetic acid, a (for acid) and the sodium acetate, b (for base). So eqn 1 is pH = pKa + log b/a
eqn 2 is a+b = 0.05
Substitute into eqn 1 as follows:
5.00 = 4.76 + log b/a
5.00-4.76 = log b/a
0.24 = log b/a
b/a = 1.74 or
b = 1.74a
eqn 2 is a + b = 0.05
Substitute b from 1 into 2.
a + 1.74a = 0.05
2.74a = 0.05
a = 0.05/2.74 = ?
Then you substitute a back into eqn 2 of a + b = 0.05. Now you know the 0.05 and a, solve for b and go from there.
pH = pKa + log (Ac^-)/(HAc)
Plug in pH, pKa, and solve for (Ac^-)/(HAc). That is equation 1.
Equation 2 is (HAc) + (Ac^-) = 0.05
Solve equations 1 and 2 simultaneously to find (Ac^-) and (HAc).
Then mol HAc = M x 0.250 = ?
and mols NaAc = M x 0.25 = ?
The above gives you the mols you will need which is what you asked for in the first part of the problem. I don't understand how the 0.1M solutions are to be used. Usually problems of this kind ask for mL of the solutions but this one doesn't do that so I ignored that part of the problem.
How do you solve those 2 equations simultaneously
kindly give me hint for this
To prepare a pH 5.00 buffer solution with a total buffer concentration of acetic acid + sodium acetate of 0.050 M, you will need to calculate the moles of acetic acid and sodium acetate required.
First, determine the desired moles of acetic acid (CH3COOH) using the formula:
moles = concentration (M) x volume (L)
Since the total volume of the buffer solution is 250.0 mL (0.250 L) and the total buffer concentration is 0.050 M, the moles of acetic acid needed can be calculated as:
moles of acetic acid = 0.050 M x 0.250 L = 0.0125 moles
Next, calculate the required moles of sodium acetate (CH3COONa). Since sodium acetate dissociates into one sodium ion (Na+) and one acetate ion (CH3COO-), the moles of sodium acetate needed will be the same as the moles of acetate ions.
Using the Henderson-Hasselbalch equation, pH = pKa + log ([A-]/[HA]), we can determine the ratio ([A-]/[HA]) needed to achieve a pH of 5.00. In this equation, [A-] represents the concentration of acetate ions (CH3COO-) and [HA] represents the concentration of acetic acid (CH3COOH).
Firstly, convert the given pH of 5.00 to pOH:
pOH = 14 - pH = 14 - 5.00 = 9.00
Since pOH = -log[OH-], we can calculate the hydroxide ion concentration ([OH-]):
[OH-] = 10^(-pOH) = 10^(-9.00) = 1.00 x 10^(-9) M
Applying the equilibrium constant expression for acetic acid dissociation, Ka = [A-][H+]/[HA], we can calculate the ratio [A-]/[HA]:
Ka = [A-][H+]/[HA]
10^(-pKa) = [A-]/[HA]
10^(-4.76) = [A-]/[HA]
Since [A-]/[HA] = 10^(-pKa), the required ratio for the buffer is 10^(-pKa), which is 10^(-4.76).
Now, considering the ratio [A-]/[HA] = 10^(-4.76), we can calculate the moles of sodium acetate needed:
moles of sodium acetate = [A-] = [HA] x [A-]/[HA]
moles of sodium acetate = moles of acetic acid x [A-]/[HA]
moles of sodium acetate = 0.0125 moles x 10^(-4.76)
To prepare the buffer, you will need approximately 0.0125 moles of acetic acid (CH3COOH) and 0.0125 moles of sodium acetate (CH3COONa) to achieve a total buffer concentration of 0.050 M.
To prepare the buffer solution:
1. Weigh out approximately 0.82 g (0.0125 moles) of acetic acid (CH3COOH)
2. Dissolve the acetic acid in enough water to make a final volume of 250.0 mL
3. Weigh out approximately 1.71 g (0.0125 moles) of sodium acetate (CH3COONa)
4. Dissolve the sodium acetate in enough water to make a final volume of 250.0 mL
5. Combine the two solutions and mix thoroughly
Note: Ensure that all the measurements are accurately made, and the chemicals are dissolved completely before combining the solutions. pH adjustments can be made if necessary by using small amounts of acids or bases such as hydrochloric acid or sodium hydroxide.