What is the pH of a buffer solution if you have 250 ml of a 1.56M Acetic Acid and you added 26.56 grams of sodium acetate (NaCH3CO2)? What is the new pH if you now add 1gram of NaOH to the buffer solution?
A.
pH = pKa + log (base)/(acid)
Convert 26.56 g NaAc to mols, convert to M, plug into the above equation and solve for pH.
B.
I like to work these by millimols (mmol) and not M.
mmols HAc = 250 x 1.56 = about 400 but you need a more accurate number.
mmols NaAc = 26.56/82 x 1000 = about 320--again you need a more accurate answer.
mmols NaOH = 1/40 x 1000 = about 25.
.........HAc + OH^- ==> Ac^- + H2O
I........400...0........320.......
add...........25...............
C........-25..-25.......+25
E........375...0........345
Substitute the E line into the HH equation and solve for the new pH. Remember those numbers above are approximations; you need to redo the entire problem using the correct numbers. Post your work if you get stuck.
To find the initial pH of the buffer solution, we first need to determine the concentration of acetic acid and sodium acetate after adding 26.56 grams of sodium acetate.
Step 1: Calculate the moles of acetic acid (CH3COOH) and sodium acetate (NaCH3CO2).
Molar mass of CH3COOH = 60.05 g/mol
Molar mass of NaCH3CO2 = 82.03 g/mol
Moles of CH3COOH = (1.56 M) x (0.250 L) = 0.390 mol
Moles of NaCH3CO2 = (26.56 g) / (82.03 g/mol) = 0.323 mol
Step 2: Calculate the total volume of the solution.
Total volume = 250 ml + the volume of NaCH3CO2 solution (assumed to be negligible)
Step 3: Calculate the concentrations of CH3COOH and CH3COO- (sodium acetate).
Concentration of CH3COOH = (0.390 mol) / (total volume in liters)
Concentration of CH3COO- = (0.323 mol) / (total volume in liters)
Step 4: Calculate the initial pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([CH3COO-] / [CH3COOH])
The pKa value for acetic acid is approximately 4.76.
Now, with a new addition of 1 gram of NaOH to the buffer solution, we can calculate the new pH.
Step 1: Calculate the moles of NaOH (sodium hydroxide).
Molar mass of NaOH = 40.00 g/mol
Moles of NaOH = (1 g) / (40.00 g/mol) = 0.025 mol
Step 2: Calculate the new concentrations of CH3COOH and CH3COO- after adding NaOH.
Since NaOH reacts with acetic acid in a 1:1 ratio, the decrease in the concentration of CH3COOH will be equal to the moles of NaOH added.
New concentration of CH3COOH = Initial concentration - moles of NaOH added
New concentration of CH3COO- = Initial concentration + moles of NaOH added
Step 3: Calculate the new pH using the Henderson-Hasselbalch equation and the updated concentrations.
pH = pKa + log ([CH3COO-] / [CH3COOH])
To calculate the pH of a buffer solution, you need to determine the concentration of the acid and its conjugate base, and use the Henderson-Hasselbalch equation.
First, let's calculate the concentration of the acid (acetic acid) and the conjugate base (sodium acetate).
1. Concentration of acetic acid:
You are given that you have 250 mL of a 1.56 M acetic acid solution. Let's convert this volume to liters:
250 mL = 250/1000 L = 0.25 L
Now, using the equation C = n/V, where C is the concentration, n is the number of moles, and V is the volume in liters, we can calculate the number of moles of acetic acid:
n = C * V
= 1.56 M * 0.25 L
= 0.39 moles
The concentration of acetic acid is 0.39 moles/L.
2. Concentration of sodium acetate:
You are given that you added 26.56 grams of sodium acetate (NaCH3CO2) to the solution. Let's calculate the number of moles:
molar mass of NaCH3CO2 = (23+12+3+16+12+16+16) g/mol
= 82 g/mol
Number of moles of sodium acetate:
n = mass / molar mass
= 26.56 g / 82 g/mol
= 0.324 moles
The concentration of sodium acetate is 0.324 moles/0.25 L = 1.296 moles/L.
Now that we have the concentrations of the acid (acetic acid) and its conjugate base (sodium acetate), we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
The pKa of acetic acid is approximately 4.76.
Calculating the pH of the buffer solution:
pH = 4.76 + log(1.296/0.39)
= 4.76 + log(3.323)
≈ 4.76 + 0.52
≈ 5.28
Therefore, the pH of the buffer solution after adding the sodium acetate is approximately 5.28.
Now, let's consider adding 1 gram of NaOH to the buffer solution:
First, calculate the number of moles of NaOH:
molar mass of NaOH = (23+16+1) g/mol
= 40 g/mol
Number of moles of NaOH:
n = mass / molar mass
= 1 g / 40 g/mol
= 0.025 moles
Since NaOH is a strong base that dissociates completely, we can assume that 0.025 moles of OH- will be added to the buffer solution.
To find the new pH, we need to determine the new concentrations of the acid and its conjugate base.
The change in concentration of acetic acid is -0.025 moles, as it reacts with the OH-.
The change in concentration of sodium acetate is +0.025 moles, as it reacts with the H+ from the dissociation of water.
Since the initial concentration of acetic acid was 0.39 moles/L and the initial concentration of sodium acetate was 1.296 moles/L, the new concentrations are as follows:
[HA] = 0.39 - 0.025 moles / 0.25 L = 1.46 moles/L
[A-] = 1.296 + 0.025 moles / 0.25 L = 5.284 moles/L
Now, we can use the Henderson-Hasselbalch equation to calculate the new pH:
pH = pKa + log([A-]/[HA])
= 4.76 + log(5.284/1.46)
= 4.76 + log(3.617)
≈ 4.76 + 0.56
≈ 5.32
Therefore, the new pH of the buffer solution after adding 1 gram of NaOH is approximately 5.32.