A solution is prepared by dissolving 50.0 g of pure HC2H3O2 and 20.0 g of NaC2H3O2 in 975 mL of solution (the final volume).
what is the ph?
What would the pH of the solution be if 50.0 mL of 0.900 M NaOH were added?
What would the pH be if 30.0 mL of 0.30 M HCl were added to the original 975 mL of buffer solution?
Please help I have tried using the Henderson Hasselbach equation and I am not getting the correct answer.
Show your work and I'll guide you through it.
First, just the pH of the solution.
I multiplied 50g by 975 mL and then 20g by 975 mL. Then I used the given pKa of 4.745 + log (19,500 / 48,750) = 4.35
Thos are molarities that go in the HH equation.
mols HAc = 50/molar mass HAc
M HAc = mols/0.975L = ?
mols NaAc = 20/molar mass NaAc.
M NaAc = mols/0.975
pH = pKa + log b/a
I get close to 4.2 but that approximate.
By the way you can take a shortcut and use mols base/mols acid and obtain the same answer as if you used M base/M acid. That's because the 975 mL cancels.
To determine the pH of the solution, we need to use the Henderson-Hasselbalch equation for a buffer solution, which is given by:
pH = pKa + log([A-]/[HA])
1. pH of the original solution:
First, we need to find out the concentrations of the acetic acid (HC2H3O2) and the acetate ion (C2H3O2-) in the solution.
(a) Concentration of HC2H3O2:
We have initially dissolved 50.0 g of pure HC2H3O2. To find the concentration (in mol/L), we need to convert grams to moles and then divide by the final volume in liters:
Molar mass of HC2H3O2 = 60.052 g/mol
Number of moles of HC2H3O2 = 50.0 g / 60.052 g/mol = 0.833 mol
Concentration of HC2H3O2 = 0.833 mol / 0.975 L = 0.854 M
(b) Concentration of NaC2H3O2:
We have initially added 20.0 g of NaC2H3O2. To find the concentration (in mol/L), we need to convert grams to moles and then divide by the final volume in liters:
Molar mass of NaC2H3O2 = 82.034 g/mol
Number of moles of NaC2H3O2 = 20.0 g / 82.034 g/mol = 0.244 mol
Concentration of NaC2H3O2 = 0.244 mol / 0.975 L = 0.251 M
(c) Calculation of pKa:
To calculate the pKa value, we need to know the Ka value for acetic acid (HC2H3O2). The Ka value for HC2H3O2 is 1.8 x 10^-5 at 25°C.
pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74
(d) Calculation of [A-]/[HA]:
[A-] = concentration of acetate ion = 0.251 M (from above)
[HA] = concentration of acetic acid = 0.854 M (from above)
Now we can substitute these values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.251/0.854)
pH ≈ 4.74 + log(0.293)
pH ≈ 4.74 - 0.532
pH ≈ 4.208
Therefore, the pH of the original solution is approximately 4.208.
2. pH after adding 50.0 mL of 0.900 M NaOH:
To find the new pH after adding NaOH, we need to consider the reaction between NaOH and HC2H3O2, which forms water and the acetate ion (C2H3O2-). The concentration of the acetate ion (C2H3O2-) will change.
(a) Number of moles of NaOH:
Molarity (M) = moles / volume (L)
Number of moles of NaOH = 0.900 M × 0.0500 L = 0.0450 mol
(b) Reaction between NaOH and HC2H3O2:
NaOH + HC2H3O2 → H2O + NaC2H3O2
Since NaOH is a strong base and HC2H3O2 is a weak acid, the NaOH will fully react with HC2H3O2.
(c) Change in the concentration of acetate ion (C2H3O2-):
- The number of moles of NaOH reacting with HC2H3O2 is equal to the number of moles of acetate ion produced.
- Since the initial concentration of NaOH is 0.900 M and the volume of NaOH added is 0.0500 L, the final concentration of C2H3O2- will be:
Final concentration of C2H3O2- = Moles of C2H3O2- / New volume (in L)
= 0.0450 mol / (0.975 L + 0.0500 L)
= 0.0450 mol / 1.025 L
= 0.0439 M
(d) Recalculation of pH using new [A-]/[HA]:
Using the new concentration of C2H3O2- and the concentration of acetic acid (0.854 M), we can recalculate the pH using the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.0439/0.854)
pH ≈ 4.74 + log(0.0514)
pH ≈ 4.74 - 1.29
pH ≈ 3.45
Therefore, the pH of the solution after adding 50.0 mL of 0.900 M NaOH is approximately 3.45.
3. pH after adding 30.0 mL of 0.30 M HCl:
To find the new pH after adding HCl, we need to consider the reaction between HCl and the acetate ion (C2H3O2-), which forms acetic acid (HC2H3O2). The concentration of acetic acid (HC2H3O2) will change.
(a) Number of moles of HCl:
Molarity (M) = moles / volume (L)
Number of moles of HCl = 0.30 M × 0.0300 L = 0.009 mol
(b) Reaction between HCl and C2H3O2-:
HCl + C2H3O2- → HC2H3O2
Since HCl is a strong acid and C2H3O2- is a weak base, the HCl will fully react with C2H3O2-.
(c) Change in the concentration of acetic acid (HC2H3O2):
- The number of moles of HCl reacting with C2H3O2- is equal to the number of moles of acetic acid produced.
- Since the initial concentration of HCl is 0.30 M and the volume of HCl added is 0.0300 L, the final concentration of HC2H3O2 will be:
Final concentration of HC2H3O2 = Moles of HC2H3O2 / New volume (in L)
= 0.009 mol / (0.975 L + 0.0300 L)
= 0.009 mol / 1.005 L
= 0.00895 M
(d) Recalculation of pH using new [A-]/[HA]:
Using the new concentration of HC2H3O2 (0.00895 M) and the concentration of acetate ion (0.0439 M), we can recalculate the pH using the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.0439/0.00895)
pH ≈ 4.74 + log(4.91)
pH ≈ 4.74 + 0.69
pH ≈ 5.43
Therefore, the pH of the solution after adding 30.0 mL of 0.30 M HCl is approximately 5.43.
I hope this explanation helps you understand the calculations involved. Let me know if you need any further clarification!