what volume of .422 M NaOH must be added to .500L of .300 M acetic acid to raise its ph to 4.00
i know you have to use the henderson hasselbalch equation but im not sure which values i would use for the concentration portion of the equation. would i have to change the molarity of any of the amounts given initially.
would i add the naoh to the acetic acid which would go to completion leaving me with 0.078 M acetic acid and .422 M of the conjugate then use those values in the equation
.177827 = (.422)(x)/(.078)(.500-x)
then x would equal the volume of NaOh needed?
thank you in advance
To solve this problem, you are on the right track by using the Henderson-Hasselbalch equation. Let's go through the steps:
1. First, write the balanced equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
2. In this reaction, acetic acid acts as a weak acid and NaOH acts as a strong base. Thus, the reaction goes to completion, and the acetic acid will be completely neutralized by the sodium hydroxide. This means that all of the acetic acid will be converted into its conjugate base, acetate (CH3COO-).
3. Now, let's consider the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH (4.00 in this case)
pKa = dissociation constant for acetic acid (4.76 for acetic acid)
[A-] = concentration of the conjugate base (acetate)
[HA] = concentration of the weak acid (acetic acid)
4. Rearrange the Henderson-Hasselbalch equation to isolate the ratio of [A-] to [HA]:
[A-]/[HA] = 10^(pH - pKa)
5. Substitute the known values:
[A-]/[HA] = 10^(4.00 - 4.76) = 10^(-0.76) = 0.177827
So, the ratio of [A-] to [HA] in the solution should be 0.177827.
6. Now, let's consider the initial concentrations given in the problem. We have 0.500 L of 0.300 M acetic acid. Using the equation:
[HA] = concentration of weak acid = 0.300 M × 0.500 L = 0.150 mol
7. Since all the acetic acid will be consumed in the reaction with NaOH, the concentration of acetic acid ([HA]) will become zero. The concentration of the conjugate base ([A-]) will be equal to the volume of sodium hydroxide added (x) multiplied by its concentration.
[A-] = concentration of conjugate base = 0.422 M × x
8. Now, substitute the known values into the ratio obtained earlier:
0.177827 = (0.422 M × x) / 0.150 mol
9. Solve the equation for x:
x = (0.177827 × 0.150 mol) / 0.422 M ≈ 0.063 L or 63 mL
Therefore, you would need to add approximately 63 mL of 0.422 M NaOH to 0.500 L of 0.300 M acetic acid to raise its pH to 4.00.
To determine the volume of 0.422 M NaOH needed to raise the pH of 0.500 L of 0.300 M acetic acid to 4.00, you can use the Henderson-Hasselbalch equation. However, the initial concentrations you mentioned should be adjusted to reflect the addition of NaOH and subsequent neutralization reaction.
Let's break down the steps:
1. Calculate the initial concentration of acetic acid (CH3COOH). The molarity (M) of acetic acid is given as 0.300 M in 0.500 L:
initial moles of CH3COOH = M × V
= 0.300 M × 0.500 L
= 0.150 moles of CH3COOH
2. Since NaOH will react with acetic acid in a 1:1 ratio to form water and sodium acetate, the initial moles of acetic acid will also determine the final moles of sodium acetate.
final moles of sodium acetate = 0.150 moles of CH3COOH
3. Calculate the final volume and concentration of the solution after adding NaOH.
final volume of the solution = 0.500 L + volume of NaOH
final moles of sodium acetate = 0.150 moles
final concentration of sodium acetate = moles / volume
= 0.150 moles / (0.500 L + volume of NaOH)
4. Use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.00 (desired pH)
pKa = 4.75 (acetic acid's pKa)
[A-] = final concentration of acetate = 0.150 moles / (0.500 L + volume of NaOH)
[HA] = final concentration of acetic acid = 0.300 M - [A-]
Substituting the values into the equation:
4.00 = 4.75 + log([0.150 / (0.500 + volume of NaOH)] / [0.300 - 0.150 / (0.500 + volume of NaOH)])
5. Rearrange and solve the equation for the volume of NaOH (x):
0.75 = log([0.150 / (0.500 + x)] / [0.150 / (0.500 + x)])
10^0.75 = [0.150 / (0.500 + x)] / [0.150 / (0.500 + x)]
10^0.75 = 1
10^0.75 = (0.150 / (0.500 + x)) / (0.150 / (0.500 + x))
(0.500 + x) = 10^0.75 × (0.500 + x)
(0.500 + x) = 3.548 × (0.500 + x)
x = 2.774
Therefore, approximately 2.774 L of 0.422 M NaOH needs to be added to raise the pH of the solution to 4.00.
Technically the concentrations go in in molarity. I like to work in mols and since M = mols/L and the value for L in both numerator and denominator is the same the L cancel and one can use mols (actually I use millimols) directly.
millimols HAc = 500 mL x 0.3M = 150 mmols.
........OH^- + HAc ==> Ac^- + H2O
I.......0......150......0......0
add.....x.......................
C......-x......-x.......x......
E.......0.....150-x.....x.......
Substitute into the HH equation and solve for x = mmols NaOH.
Then since M = mmols/mL, plug in M NaOH and mmols NaOH to find mL 0.422 M NaOH. I think the answer is approx 50 mL.
I....