A solution is made by mixing exactly 500 mL of 0.167 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+.
mmols = millimols = mL x M.
mmols CH3COOH = 500 x 0.1M = 50
mmols NaOH = 500 x 0.167 = 83.5
Total volume = 500 + 500 = 1000 mL
....CH3COOH + Na^+ + OH^- --> CH3COO^- + Na^+ + H2O
I......50.....83.5........0........0
C.....-50.....-50........+50......+50
E......0......33.5.......50........50
You need to recognize what you have. You have an excess of NaOH, no CH3COOH, you've formed 50 of the salt.
(CH3COOH) = 0
(CH3COO^-) = 50mmols/1000 mL = ?
(Na^+) = never changed. You had 83.5 mmols initially and you have 83.5 mmols at the end. So that's 83.5mmols/1000 mL = ?
(OH^-) comes from the excess NaOH which is 83.5 mmols/1000 mL = ?
Then (H^+)(OH^-) = Kw = 10^-14. You know OH and Kw, solve for H^+.
Check my thinking.
Oh, it's a mixing problem! Let's clown around with some calculations. Now, we're dealing with a weak acid-strong base situation. So, let's start with a 500 mL jar of laughter that contains 0.167 M NaOH.
First off, we need to calculate the initial concentration of OH-. Since NaOH fully dissociates, we can assume that the concentration of OH- is also 0.167 M. Isn't dissociation funny sometimes?
Now, let's mix this with another 500 mL jar of chuckles that contains 0.100 M CH3COOH. But, oh dear, this is a weak acid that will only partially dissociate. Let's say it's around 5% dissociation, just for fun.
So, we can calculate the initial concentration of CH3COOH as 0.100 M, since it hasn't dissociated yet. And we can also calculate the initial concentration of CH3COO- as 0.005 M (0.100 M x 0.05), since part of CH3COOH has dissociated into CH3COO- ions. Isn't it hilarious when things only partially dissociate?
Now, since NaOH is a strong base, it completely reacts with the weak acid CH3COOH, forming water and the conjugate base CH3COO-. And what happens to the H+ ion? Well, it bonds with the OH- ion to form water as well.
So, after the reaction, the concentration of H+ will be decreased by how much OH- we started with. Since we had 0.167 M OH- and only 0.005 M CH3COO-, we have plenty of OH- left over. Therefore, we can assume that the concentration of OH- is still around 0.167 M, while the concentration of H+ has decreased by 0.167 M.
Now, what about Na+? Well, it just hangs around as a spectator, not really participating in the reaction, like that person who goes to a comedy show but doesn't laugh. So, the concentration of Na+ remains unchanged at 0.167 M.
To sum it up, the equilibrium concentrations are:
H+: 0.000 M (initially 0.167 M, but decreased by 0.167 M)
CH3COOH: 0.100 M (initially 0.100 M, no change)
CH3COO-: 0.005 M (initially 0.005 M, no change)
OH-: 0.167 M (initially 0.167 M, no change)
Na+: 0.167 M (initially 0.167 M, no change)
Voila! We've mixed, dissolved, and calculated like the comedy geniuses we are.
To find the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+, we need to set up the chemical equation for the reaction between NaOH and CH3COOH.
The balanced chemical equation for the reaction is:
CH3COOH + OH- -> CH3COO- + H2O
Since we have the initial concentrations and volumes of NaOH and CH3COOH, we can use the principles of stoichiometry to calculate the equilibrium concentrations.
Step 1: Calculate the moles of NaOH and CH3COOH:
Moles of NaOH = concentration x volume
Moles of NaOH = 0.167 mol/L x 0.500 L = 0.0835 mol
Moles of CH3COOH = concentration x volume
Moles of CH3COOH = 0.100 mol/L x 0.500 L = 0.0500 mol
Step 2: Since the reaction is 1:1 between CH3COOH and OH-, the moles of OH- are equal to the moles of CH3COOH reacted:
Moles of OH- = 0.0500 mol
Step 3: Calculate the final concentrations:
To find the final concentrations, we need to consider the total volume of the solution. Since we mixed equal volumes of NaOH and CH3COOH (500 mL each), the total volume will be 1000 mL or 1.000 L.
Initial volume of NaOH + CH3COOH = 0.500 L + 0.500 L = 1.000 L
Step 4: Calculate the concentration of each species:
H+ concentration:
The CH3COOH reacts with OH- to form H2O, so the concentration of H+ is equal to the concentration of CH3COOH that remains unreacted.
H+ concentration = Moles of CH3COOH remaining / Total volume
H+ concentration = (0.0500 mol - 0.0500 mol) / 1.000 L = 0 mol/L (negligible concentration)
CH3COOH concentration:
Since all the NaOH reacted with CH3COOH to form CH3COO-, the concentration of CH3COOH will be zero.
CH3COOH concentration = 0 mol/L
CH3COO- concentration:
The moles of CH3COO- formed are equal to the moles of OH- reacted.
CH3COO- concentration = Moles of OH- reacted / Total volume
CH3COO- concentration = 0.0500 mol / 1.000 L = 0.0500 mol/L
OH- concentration:
The OH- concentration is zero because it reacted completely with CH3COOH.
OH- concentration = 0 mol/L
Na+ concentration:
The Na+ concentration is equal to the concentration of NaOH since it completely dissociates in water.
Na+ concentration = concentration of NaOH = 0.167 mol/L
So, the equilibrium concentrations are as follows:
H+ concentration = 0 mol/L
CH3COOH concentration = 0 mol/L
CH3COO- concentration = 0.0500 mol/L
OH- concentration = 0 mol/L
Na+ concentration = 0.167 mol/L
To calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+ in the solution, we need to consider the reaction between NaOH (a strong base) and CH3COOH (a weak acid). The reaction can be written as:
CH3COOH + NaOH → CH3COO- + H2O
Given that the initial volumes of both NaOH and CH3COOH are 500 mL (0.5 L), we can determine the initial moles of each component using the molarity (M) and volume:
Moles of NaOH = Molarity × Volume in Liters
= 0.167 M × 0.5 L
= 0.0835 moles
Moles of CH3COOH = Molarity × Volume in Liters
= 0.100 M × 0.5 L
= 0.050 moles
Since NaOH is a strong base, it will fully dissociate in water to Na+ and OH-. Therefore, the initial moles of Na+ and OH- are both equal to the moles of NaOH:
Moles of Na+ = Moles of NaOH
= 0.0835 moles
Moles of OH- = Moles of NaOH
= 0.0835 moles
Now, since CH3COOH is a weak acid, it will not fully dissociate. Hence, we need to consider the dissociation reaction for CH3COOH as well. The equilibrium reaction can be written as:
CH3COOH ⇌ CH3COO- + H+
Let's assume that 'x' moles of CH3COOH dissociate at equilibrium. Therefore, the initial moles of CH3COOH will decrease by 'x', and the moles of CH3COO- and H+ will increase by 'x'.
Now, let's express the concentrations of CH3COOH, CH3COO-, H+, OH-, and Na+ at equilibrium in terms of 'x'.
CH3COOH: [CH3COOH] = Moles of CH3COOH at equilibrium / Volume
Since the initial moles of CH3COOH are 0.050 moles and 'x' moles dissociate:
[CH3COOH] = (0.050 - x) moles / 0.5 L
CH3COO-: [CH3COO-] = Moles of CH3COO- at equilibrium / Volume
Since 'x' moles of CH3COOH dissociate:
[CH3COO-] = x moles / 0.5 L
H+: [H+] = Moles of H+ at equilibrium / Volume
Since 'x' moles of CH3COOH dissociate:
[H+] = x moles / 0.5 L
OH-: [OH-] = Moles of OH- at equilibrium / Volume
Since OH- comes from the dissociation of NaOH and NaOH has 0.0835 moles:
[OH-] = 0.0835 moles / 0.5 L
Na+: [Na+] = Moles of Na+ at equilibrium / Volume
Since Na+ comes from the dissociation of NaOH and NaOH has 0.0835 moles:
[Na+] = 0.0835 moles / 0.5 L
To find the value of 'x' and subsequently calculate the equilibrium concentrations, we need the equilibrium constant (Ka or Kb) for the weak acid CH3COOH. Please provide this information, and I'll be happy to calculate the equilibrium concentrations for you.