A solution is made by mixing exactly 500 mL of 0.178 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+.
The secret to solving these problems is to recognize what you have in solution at the point which the question addresses. Then write an equation and prepare an ICE table. Like the following which I will start but you must finish.
500 mL x 0.178 = 89.0 mmoles NaOH
500 mL x 0.100 = 50.0 mmoles CH3COOH
.....NaOH + CH3COOH ==> CH3COONa + H2O
initial 89.0..50.0........0.........0
change..-50.0.-50.0......+50.0..+50.0
equil...39.0...0..........50.0..50.0
So you have 39.0 mmoles NaOH after the reaction; therefore, the pH will be determined by the OH^- from NaOH.
(OH^-) = mmoles/mL
mmoles = 39.0 and mL = 1000 mL
(OH^-) = 0.039mols/L = 0.0390 M
The others are done similarly.
I calculated all of them and this is what I got:
1) [H+] = 2.56x10^-13 M
2) [OH-] = .039 M
3) [Na+] .089 M
4) [CH3COO-] = .05
5) [CH3COOH] = 0
All of them were right except number 5 and I don't know what the answer to number 5 is suppose to be.
I would have agreed with all of your answers, including number 5, because whatever it is it must be so small as to be negligible. YOu might try this.
CH3COOH ==> H^+ + CH3COO^-
Ka = (H^+)(CH3COO^-)/(CH3COOH)
If I substitute Ka = 1.8 x 10^-5 (your text may have a different number) and (H^+) from the problem as 2.56E-13 along with (CH3COO^-) = 0.05 from the problem and solve for (CH3COOH), I get something like 7.11E-10 which is, indeed, quite small. Or we could look at it another way, CH3COO^- is a base and hydrolyzes to produce
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (Kw/Ka) = (1E-14/1.8E-5) = 5.55E-10 = (CH3COOH)(OH^-)/(CH3COO^-).
If I substitute 0.039 for OH and 0.05 for CH3COO^- and solve for CH3COOH, I get 7.11E-10. Same value. That's a reasonable answer I think and it IS small. Let me know if this doesn't do it.
yes 7.11E-10 was the right answer thank you very much for the help :)
Wait, then why isn't the concentration of [H+] also considered to be zero since the value is so small?
To calculate the equilibrium concentrations of the various ions in the solution, we can use the principles of acid-base chemistry and equilibrium calculations.
First, let's start by writing the balanced chemical equation for the reaction between NaOH and CH3COOH:
CH3COOH + NaOH -> CH3COO- + Na+ + H2O
We know that NaOH is a strong base and completely ionizes in water, while CH3COOH is a weak acid and only partially ionizes. Therefore, we can write the dissociation reactions as follows:
NaOH dissociation:
NaOH -> Na+ + OH-
CH3COOH dissociation:
CH3COOH -> CH3COO- + H+
Since NaOH dissociates completely, the concentration of OH- ions will be the same as the initial concentration of NaOH, which is 0.178 M.
Now, to calculate the equilibrium concentrations, we need to use the concept of stoichiometry. Since we have equal initial volumes of NaOH and CH3COOH solutions, the number of moles of each species will be the same after the reaction has reached equilibrium. Let x be the concentration (in M) of CH3COO- and H+ ions after the reaction.
Therefore, the equilibrium concentrations will be as follows:
H+ concentration: x
CH3COOH concentration: 0.100 M - x
CH3COO- concentration: x
OH- concentration: 0.178 M
Na+ concentration: 0.178 M
Note that the initial concentration of CH3COOH (0.100 M) is equal to the concentration of the remaining undissociated CH3COOH at equilibrium (0.100 M - x). This is assuming that the extent of dissociation of CH3COOH is small compared to its initial concentration.
To find the value of x, we need to use the equilibrium constant expression for the dissociation of CH3COOH:
Ka = [CH3COO-][H+] / [CH3COOH]
The value of Ka for acetic acid is 1.8 x 10^-5 at 25°C. Since the initial concentration of CH3COOH is 0.100 M, and assuming x is small compared to 0.100 M, we can approximate the equilibrium concentration of CH3COOH as 0.100 M.
Now, let's plug in the known values to solve for x:
1.8 x 10^-5 = x * x / (0.100 - x)
Simplifying and solving the quadratic equation will give us the value of x, which can then be used to calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+.
Note: The quadratic equation may be solved using numerical methods, such as the Newton-Raphson method, or online solvers.
Once we have the value of x, we can substitute it back into the equilibrium concentration expressions to calculate the exact values of each ion's concentration.
It's important to note that this process assumes that the volume of the solution remains constant and that there are no other reactions taking place that might influence the equilibrium concentrations.