To 400mL of a 0.1 M CH3COOH solution(Ka=1.8 x 10-5),two subsequent additions of 100mL of 0.2 M NaOH solution have been done.determine the solution pH after each addition.
Drbob,I don't understand this problem how to solve.Can you help me and give me the full calculation.
First, please spell chemistry right. I'm annoyed. Second, this is a buffer problem for the first addition. They throw you a curve with the second addition because it is NOT a second buffer problem. Here is the scoop.
Call acetic acid, CH3COOH, as HAc.
millimols HAc = mL x M = 400 x 0.1 = 40.
mmols NaOH (first time) = mL x M = 20
mmols NaOH (second time) = 20
Ka = 1.8E-5; pKa = -log Ka = 4.74
Henderson-Hasselbalch equation is
pH = pKa + log (base)/(acid) and since the volumes are the same for each separate problem we can use millimoles and not actual concentrations (at least for the first addition).
.......HAc + OH^- ==> Ac^- + H2O
I......40....0.........0......0
add.........20.................
C.....-20...-20.......20......
E......20....0........20......
Then use the HH equation, plug the E line into the HH equation and solve for pH. The answer is 4.74.
For the second addition of 20 mmols NaOH we do this and START with the end of the first one.
.......HAc + OH^- ==> Ac^- + H2O
I......20....0........20......
add..........20................
C.....-20...-20.......20......
E.......0....0........40
And here is the catch. Notice that the second addition of NaOH has neutralized ALL of the HAc and there is no acid left so this is no longer a buffer problem. What do we have in the solution? Just 40 millimols Ac^- (NaAc or sodium acetate) in 400 mL + 100 mL + 100 mL H2O so (Ac^-) = mmols/mL = 40/600 = 0.0667M NaAc. So the pH is determined by the hydrolysis of the NaAc, like this.
........Ac^- + HOH => HAc + OH^-
I.....0.0667...........0......0
C.......-x............x......x
E.....0.0667-x.........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.0667-x) and solve for x = (OH^-), then convert to pH. I didn't work this particular problem but most acetate hydrolysis problems come out to be in the pH range of 8.3 to 9. With a starting concn of 0.0667 sometimes you must use the quadratic equation (that is, 0.0667-x must be used and you can't ignore the -x) so make sure the quadratic is not required.
Post your work if you run into trouble with this and explain in detail what you don't understand about it.
A quadratic is not required and you can let 0.0667-x = 0.0667.
Sure! I can help you solve this problem step by step.
To find the solution's pH after each addition, we need to consider the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
Here's what happens in each addition:
1) First Addition: 100mL of 0.2 M NaOH
a) Calculation to determine the moles of CH3COOH initially present:
- Moles of CH3COOH = volume (L) × concentration (mol/L)
- Volume of CH3COOH = 400 mL = 0.4 L
- Concentration of CH3COOH = 0.1 M
- Moles of CH3COOH = 0.4 L × 0.1 mol/L = 0.04 mol
b) Calculation to determine the moles of NaOH added:
- Moles of NaOH = volume (L) × concentration (mol/L)
- Volume of NaOH = 100 mL = 0.1 L
- Concentration of NaOH = 0.2 M
- Moles of NaOH = 0.1 L × 0.2 mol/L = 0.02 mol
c) Now, let's determine which reactant will limit the reaction:
- The balanced chemical equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
- According to the equation, one mole of CH3COOH reacts with one mole of NaOH.
Since the number of moles of NaOH (0.02 mol) added is less than the number of moles of CH3COOH (0.04 mol) initially present, NaOH is the limiting reactant.
d) Calculation to determine the moles of CH3COOH remaining:
- Moles of CH3COOH remaining = Moles of CH3COOH initially present - Moles of NaOH added
- Moles of CH3COOH left = 0.04 mol - 0.02 mol = 0.02 mol
e) Calculation to determine the new concentration of CH3COOH:
- Volume of CH3COOH = 400 mL = 0.4 L (unchanged)
- New concentration of CH3COOH = moles of CH3COOH remaining / volume of CH3COOH
- New concentration of CH3COOH = (0.02 mol) / (0.4 L) = 0.05 M
f) Calculation to determine the concentration of CH3COONa formed:
- From the balanced equation, we see that the mole ratio between CH3COOH and CH3COONa is 1:1.
- Since both reactants have equal moles, the concentration of CH3COONa formed is also 0.05 M.
Now, to determine the solution pH, we need to calculate the concentration of H+ ions. Since acetic acid is a weak acid, we will consider the dissociation of acetic acid using the equilibrium expression and the acidity constant (Ka = 1.8 × 10^(-5)).
The dissociation of CH3COOH is given by:
CH3COOH ⇌ CH3COO- + H+
To find the concentration of H+ ions, we need to use the equilibrium constant expression:
Ka = [CH3COO-][H+]/[CH3COOH]
g) Calculation to determine the concentration of H+ ions:
- We know that [CH3COO-] = concentration of CH3COONa formed = 0.05 M (from above)
- We determined that the concentration of CH3COOH remaining is also 0.05 M (from above)
- Let's assign the concentration of H+ ions as "x"
Substituting into the equilibrium constant expression:
Ka = (0.05 M) * x / (0.05 M)
Now, let's solve for "x":
Ka = x
Since x = Ka, the concentration of H+ ions is equal to the Ka value, which is 1.8 × 10^(-5) M.
h) Calculation to determine the pH:
pH = -log[H+]
pH = -log(1.8 × 10^(-5))
Using a calculator, we find that the pH after the first addition is approximately 4.74.
Now, you can continue with the second addition of NaOH. Let me know if you need help with that as well.