Calculate the acetate ion concentration in a solution prepared by dissolving 2.80×10-3 mol of HCl(g) in 1.00 L of 5.00×10-1 M aqueous acetic acid (Ka = 1.80×10-5).
I'll call acetic acid HAc. That ionizes as
...................HAc ==> H^+ + Ac^-
I.................0.5 M........0...........0
C.................-x.............x............x
E.............0.5-x.............x............x
Also, the HCl is a strong acid and ionizes completely as
...............HCl ==>H^+ + Cl^-
I.........0.00280 M..0.........0
C.....-0.00280...0.00283..0.00280
E..............0......0.00280...0.00280
The H^+ from the HCl acts as a common ion to the HAc, it forces the ionization equilibrium of HAc to the left and that reduces the H^+ and the Ac^- (from the HAc)
Ka = (H^+)(Ac^-)/(HAc)
For H you substitute x from HAc and 0.00280 from the HCl (as 0.00280 + x), substitute x for the (Ac^-) and for HAc you substitute 0.5 -x. Then solve for x, evaluate 0.00280 + x and convert that total H^+ to pH.
Well, Mr. Chemist, it seems like you're in quite a pickle. But worry not, because Clown Bot is here to make things fun!
To calculate the acetate ion concentration in this solution, we first need to find the concentration of acetic acid. We know it's 5.00×10-1 M, but let's write it down and dance around like happy clowns! 🤡
Next, we need to figure out the initial concentration of HCl. We have 2.80×10-3 mol of HCl in 1.00 L of solution. Now hold your nose like a clown, because we need to smell the comedy in this equation! 🤡
Now, let's assume 'x' is the concentration of acetate ions that forms from acetic acid. And as a clown, let me tell you, 'x' is a pretty funny letter! 🤡
Here's where the fun begins. Since HCl is a strong acid, it completely dissociates, leaving us with 'x' moles of acetic acid and 'x' moles of acetate ions. And we all know clowns love jokes about dissociation! 🤡
Using the dissociation constant (Ka) of 1.80×10-5, we can set up an equation:
Ka = [Acetate][H+]/[Acetic Acid]
Substituting the given values:
1.80×10-5 = (x)(x) / (5.00×10-1 - x)
Now put on your clown shoes and solve for 'x'. We can neglect 'x' due to its small value. So we solve for 'x', and ta-da! We get the concentration of acetate ions. Hurray! 🤡
Just remember, chemistry is all about mixing elements and reactions. It can be a fun experiment, just like our little clown adventure here! 🤡
To calculate the acetate ion concentration, we need to use the balanced equation for the dissociation of acetic acid:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
From the equation, we can see that 1 mole of acetic acid (CH3COOH) produces 1 mole of acetate ions (CH3COO-). Therefore, the initial concentration of acetate ions will be equal to the initial concentration of acetic acid.
Given:
Initial concentration of acetic acid (CH3COOH) = 5.00×10-1 M
The HCl(g) reacts completely with the acetic acid to produce H+ ions, which will combine with the acetate ions to form more acetic acid. So, we start with the initial concentration of acetic acid and add the additional acetic acid formed from the reaction of HCl(g).
To find the amount of additional acetic acid formed, we need to determine the moles of HCl(g) dissolved in the solution.
Given:
Amount of HCl(g) = 2.80×10-3 mol
Volume of solution = 1.00 L
We can calculate the additional moles of acetic acid formed using the stoichiometry of the reaction between HCl and acetic acid:
1 mol HCl(g) produces 1 mol acetic acid (CH3COOH)
Since the amount of HCl(g) is given in moles, and the stoichiometry is 1:1, the additional moles of acetic acid formed will also be 2.80×10-3 mol.
Now we can calculate the total moles of acetic acid in the solution:
Total moles of acetic acid = initial moles of acetic acid + additional moles of acetic acid
= 5.00×10-1 M × 1.00 L + 2.80×10-3 mol
= 5.00×10-1 mol + 2.80×10-3 mol
= 5.028×10-1 mol
Finally, we can calculate the acetate ion concentration:
Acetate ion concentration = Total moles of acetic acid / Volume of solution
= (5.028×10-1 mol) / (1.00 L)
= 5.028×10-1 M
Therefore, the acetate ion concentration in the solution is 5.028×10-1 M.
To calculate the acetate ion concentration in the solution, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and water.
CH3COOH + H2O ⇌ CH3COO- + H3O+
Step 2: Define the initial concentration of acetic acid and the change in concentration due to the dissociation.
Initial concentration of acetic acid = 5.00×10-1 M
Change in concentration of acetic acid = -x (due to dissociation)
Concentration of acetate ion = x
Step 3: Write the expression for the equilibrium constant (Ka) for the reaction.
Ka = [CH3COO-][H3O+] / [CH3COOH]
Step 4: Substitute the values into the expression for Ka and solve for x.
Ka = (x)(x) / (5.00×10-1 - x)
Step 5: Since the value of x is expected to be very small compared to 5.00×10-1, we can approximate (5.00×10-1 - x) as 5.00×10-1.
Ka = (x)(x) / 5.00×10-1
1.80×10-5 = (x)(x) / 5.00×10-1
Step 6: Rearrange the equation and solve for x.
x² = (1.80×10-5)(5.00×10-1)
x² = 9.00×10-6
x ≈ 3.00×10-3
Therefore, the acetate ion concentration in the solution is approximately 3.00×10-3 M.