Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

Solution A:
10.0 mL 3.0e-4M bromescol green solution
25.0mL 1.60M acetic acid (HAc)
10.0mL .200M KCl solution

Solution B: 10.0mL 3.0e-4M bromescol green solution
10.0mL of .160M sodiuma cetate solution

a.Calculate moles Ac- in solution B
b. Calculate the molarity of HAc in solution A.
c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
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K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
a. Calculate pKa
b. calculate moles HAc in 5mL solution A added to solution B
c.Calculate moles Ac- in solution B
d. Rearrange equation to solve for pH

Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

Solution A:
10.0 mL 3.0e-4M bromescol green solution
25.0mL 1.60M acetic acid (HAc)
10.0mL .200M KCl solution

Solution B: 10.0mL 3.0e-4M bromescol green solution
10.0mL of .160M sodiuma cetate solution

The following three seem clear enough.
a.Calculate moles Ac- in solution B
(Note: mols = M x L = ? but I like to work in millimoles). So 10 mL x 0.160M = 1.6 mmols = 0.0016 mols.

b. Calculate the molarity of HAc in solution A.
1.6M x ((25 mL of HAc/45 mL total) = 0.888M HAc.

c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
mLA x MA = mmols B
mLA x 0.8889M = 1.6
mL A = 1.6/0.8889 = about 1.799 which I would round to 1.8 mL. You see why I like to work in mmols. mL x M = mmols AND M = mmols/mL.)


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K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
a. Calculate pKa
pKa = -log Ka = -log 2E-5 = about 4.70 but you need to confirm that and round to the right number of s.f.

b. calculate moles HAc in 5mL solution A added to solution B
mL A + M B = mmols HAc in B = 5 mL x 0.8889 = 4.45 millimols = 0.00445 mols HAc.

c.Calculate moles Ac- in solution B
Isn' that done above in the first part a? 0.0016 mols Ac^- (or 1.6 mmols).

d. Rearrange equation to solve for pH
The rearranged equation is pH = pKa + log(base)/(acid)
pH = 4.70 + log (1.60/4.45) = ? or if you want to use mols it is
pH = 4.80 + log(0.0016/0.00445) = ?

Check all of this carefully. Post any questions but show your work and explain exactly what your trouble is.

The molarity of HAc in solution A is wrong according to my TA.

a. Sure, let's first calculate the moles of Ac- in solution B. We are given the concentration of the sodium acetate-indicator solution, which is 3.0e-4 M. To find the moles, we can multiply the concentration by the volume of the solution:

Moles of Ac- in solution B = (3.0e-4 M) * (10.0 mL)
= 3.0e-3 moles

b. Next, let's calculate the molarity of HAc in solution A. We are given the volume and concentration of acetic acid (HAc) in solution A:

Molarity of HAc in solution A = (1.60 M) * (25.0 mL / 1000 mL)
= 0.04 M

c. To calculate the volume of solution A needed when the moles of Ac- equals moles of HAc, we can use the equation:
Moles HAc = Moles Ac-
(0.04 M) * (Volume of solution A in mL / 1000 mL) = 3.0e-3 moles

Rearranging the equation, we find:
Volume of solution A = (3.0e-3 moles * 1000 mL) / (0.04 M)
= 75 mL

So, the volume of solution A needed to obtain a buffer solution that is equimolar in acetate and acetic acid is 75 mL.

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a. Let's calculate pKa using the given equation: pKa = pH + log(N HAc/ N Ac-)

b. We are adding 5 mL of solution A to solution B. To calculate the moles of HAc in this added volume, we can use the concentration and volume:

Moles of HAc in 5 mL solution A = (0.04 M) * (5.0 mL / 1000 mL)
= 0.0002 moles

c. Since we know that moles of HAc equal moles of Ac-, the moles of Ac- in solution B is also 0.0002 moles.

d. To rearrange the equation pKa = pH + log(N HAc/ N Ac-) to solve for pH, we can subtract log(N HAc/ N Ac-) from both sides:

pKa - log(N HAc/ N Ac-) = pH

So, rearranging the equation, we can solve for pH.

a. To calculate the moles of Ac- in solution B, we need to use the molarity and volume of sodium acetate.

Moles of Ac- = Molarity * Volume
Moles of Ac- = (3.0e-4 M) * (10.0 mL)
Moles of Ac- = 3.0e-6 moles

Therefore, there are 3.0e-6 moles of Ac- in solution B.

b. To calculate the molarity of HAc in solution A, we need to use the moles of acetic acid and the volume of solution A.

Molarity of HAc = Moles of HAc / Volume
Molarity of HAc = (1.60 M) * (25.0 mL) / (10.0 mL)
Molarity of HAc = 4.0 M

Therefore, the molarity of HAc in solution A is 4.0 M.

c. To calculate the volume of solution A needed to obtain an equimolar buffer, we need to set the moles of HAc equal to the moles of Ac- and solve for the volume of solution A.

Moles of HAc = Moles of Ac-
Molarity of HAc * Volume of solution A = Molarity of Ac- * Volume of solution B

(4.0 M) * (Volume of solution A) = (3.0e-6 moles) * (10.0 mL)
Volume of solution A = (3.0e-6 moles * 10.0 mL) / (4.0 M)
Volume of solution A = 7.5e-6 L = 7.5 mL

Therefore, 7.5 mL of solution A must be added to solution B to obtain an equimolar buffer.

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a. To calculate pKa, we need to use the dissociation constant equation and the pH.

pKa = pH + log(N HAc / N Ac-)

pKa = pH + log((Molarity of HAc * Volume of solution A) / (Molarity of Ac- * Volume of solution B))

b. To calculate the moles of HAc in 5 mL of solution A added to solution B, we can use the molarity and volume.

Moles of HAc = Molarity * Volume
Moles of HAc = (4.0 M) * (5.0 mL)
Moles of HAc = 2.0e-5 moles

c. To calculate the moles of Ac- in solution B, we already have the moles of Ac- from part a.

Moles of Ac- = 3.0e-6 moles

d. To rearrange the equation to solve for pH, we use the definition of logarithm.

pKa - pH = log((Molarity of HAc * Volume of solution A) / (Molarity of Ac- * Volume of solution B))

pH = pKa - log((Molarity of HAc * Volume of solution A) / (Molarity of Ac- * Volume of solution B))

Plug in the known values and calculate pH.

To solve this problem, we need to follow a series of steps:

a. To calculate the moles of Ac- (acetate ion) in solution B, we need to use the formula:
moles Ac- = volume (in liters) * molarity
Given that the volume of solution B is 10.0 mL and the molarity of bromescol green solution (used as an indicator) is 3.0e-4 M, we have:
moles Ac- = 10.0 mL * (3.0e-4 M / 1000 mL/1 L) = 3.0e-6 moles Ac-

b. To calculate the molarity of HAc (acetic acid) in solution A, we can use the equation:
Molarity = moles / volume (in liters)
In solution A, the volume of acetic acid is given as 25.0 mL, which is equivalent to 0.025 L. Since the concentration of acetic acid is 1.60 M, we have:
Molarity HAc = (1.60 mol/L) * (0.025 L) = 0.04 moles HAc

c. To determine the volume of solution A needed for the moles of Ac- to equal the moles of HAc, we can set up an equation using the molarity and volumes of the two solutions. Let's assume V is the volume of solution A required.
Moles HAc = Moles Ac-
Molarity HAc * V = Molarity Ac- * (10.0 mL /1000 mL / 1 L)
Substituting the values we know, we have:
0.04*V = 3.0e-6*(10.0/1000/1)
0.04*V = 3.0e-7
V = (3.0e-7) / (0.04)
V = 7.5e-6 L or 7.5 mL

Therefore, 7.5 mL of solution A must be added to solution B to obtain a buffer solution that is equimolar in acetate and acetic acid.

Now let's move on to the second part of the question:

a. The equation pKa = pH + log(N HAc/ N Ac-) can be rearranged to calculate pKa. Rearranging this equation, we have:
pKa - pH = log(N HAc/ N Ac-)
By taking the inverse log (10^x), we get:
10^(pKa - pH) = N HAc/ N Ac-
Taking the log of the ratio of concentrations of HAc and Ac- will give us pKa.

b. To calculate the moles of HAc in 5 mL of solution A added to solution B, we need to use the formula:
moles HAc = molarity * volume (in liters)
Given that the molarity of HAc is 0.04 M and the volume is 5 mL (0.005 L), we have:
moles HAc = 0.04 mol/L * 0.005 L = 2.0e-4 moles HAc

c. To calculate the moles of Ac- in solution B, we can use the formula:
moles Ac- = volume (in liters) * molarity
Given that the volume of solution B is 10.0 mL, which is equivalent to 0.010 L, and the molarity of bromescol green is 3.0e-4 M, we have:
moles Ac- = 0.010 L * (3.0e-4 M / 1 L) = 3.0e-6 moles Ac-

d. To solve for pH, we can rearrange the equation pKa = pH + log(N HAc/ N Ac-) as:
pH = pKa - log(N HAc/ N Ac-)
Substituting the values we know, we have:
pH = pKa - log(2.0e-4 / 3.0e-6)
Calculating log(2.0e-4 / 3.0e-6) and subtracting it from the given pKa value will give us pH.