Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution: a solution that is 0.170 M in HC2H3O2 and 0.105 M in KC2H3O2

I agree that the Henderson-Hasslebalch equation is easier, but how do you find the pka without doing an ice table first?

pKa = -log(Ka)

You don't need an ICE table to do this.

Use the Henderson-Hasselbalch equation.

Sure! Let's solve this equilibrium problem and calculate the pH of the solution. We can start by writing the balanced equation for the dissociation of HC2H3O2:

HC2H3O2 (aq) ⇌ H+ (aq) + C2H3O2- (aq)

We can assume that the concentration of H+ is equal to the concentration of HC2H3O2 that dissociates and the concentration of C2H3O2- is equal to the initial concentration of KC2H3O2.

Now, let's create an ICE table:

HC2H3O2 (aq) ⇌ H+ (aq) + C2H3O2- (aq)
Initial: 0.170 M 0 M 0.105 M
Change: -x M +x M +x M
Equilibrium: 0.170 - x M x M 0.105 + x M

Using the equilibrium expression for Ka of HC2H3O2: Ka = [H+][C2H3O2-]/[HC2H3O2], we can set up the equation:

Ka = (x)(0.105 + x)/(0.170 - x)

Since the value of Ka for HC2H3O2 is 1.8 x 10^-5, we can substitute it in and solve for x.

1.8 x 10^-5 = (x)(0.105 + x)/(0.170 - x)

And, x << 0.170 and 0.105, so we can ignore the "-x" in the denominator.

1.8 x 10^-5 = (x)(0.105 + x)/0.170

0.170(1.8 x 10^-5) = x(0.105 + x)

0.00306 = 0.105x + x^2

x^2 + 0.105x - 0.00306 = 0

Using the quadratic formula, we get:

x = (-0.105 ± √(0.105^2 - 4(1)(-0.00306)))/2(1)

Simplifying that equation, we find:

x = (-0.105 ± 0.146)/2

Considering only the positive root, x ≈ 0.021

Now, we can find the concentration of H+ ions, which is approximately equal to 0.021 M.

Finally, we can calculate the pH using the formula pH = -log[H+]:

pH = -log(0.021) ≈ 1.68

So, the pH of the solution that is 0.170 M in HC2H3O2 and 0.105 M in KC2H3O2 is approximately 1.68.

To calculate the pH of a solution using an ICE table, we need to set up and solve an equilibrium expression. Let's start by writing the balanced chemical equation for the dissociation of HC2H3O2 (acetic acid) in water:

HC2H3O2 (aq) ⇌ H+ (aq) + C2H3O2- (aq)

The initial concentrations of HC2H3O2 and KC2H3O2 are given as 0.170 M and 0.105 M, respectively. Let's assume x is the concentration of H+ ions formed when HC2H3O2 dissociates.

To set up the ICE table, we go through the following steps:

1. Write the initial concentrations.
HC2H3O2 (aq): 0.170 M
H+ (aq): 0 M
C2H3O2- (aq): 0 M

2. Calculate the change in concentration.
HC2H3O2 (aq): -x
H+ (aq): +x
C2H3O2- (aq): +x

3. Write the equilibrium concentrations.
HC2H3O2 (aq): 0.170 - x
H+ (aq): x
C2H3O2- (aq): x

Now, we can write the equilibrium expression for this dissociation:

K = [H+][C2H3O2-] / [HC2H3O2]

The equilibrium constant, K, for acetic acid is approximately 1.8 x 10^(-5) at 25°C. Substituting the values into the expression:

1.8 x 10^(-5) = (x)(x) / (0.170 - x)

Since the value of x is expected to be small compared to the initial concentration of HC2H3O2, we can assume that (0.170 - x) is approximately 0.170.

Now, we can solve for x:

1.8 x 10^(-5) = (x^2) / 0.170

Rearranging the equation:

x^2 = 1.8 x 10^(-5) * 0.170

x^2 = 3.06 x 10^(-6)

Taking the square root of both sides:

x ≈ 1.75 x 10^(-3) M

This is the concentration of H+ ions in the solution. To find the pH, we take the negative logarithm (base 10) of the H+ concentration:

pH = -log[H+]

pH = -log(1.75 x 10^(-3))

pH ≈ 2.76

Therefore, the pH of the solution containing 0.170 M HC2H3O2 and 0.105 M KC2H3O2 is approximately 2.76.

The book says to use ICE

OK. The book wants you to do it the hard way.

Here are the two equilibria.
......CH3COOH ==> H^+ + CH3COO^- for which
I.....0.170.......0......0
C......-x.........+x.....+x
E.....0.170-x.......x.......x

..........CH3COOK ==> K^+ + CH3COO^-
I.........0.105.......0.......0
C.........-0.105.....0.105...0.105
E.........0..........0.105.....0.105

Now write the Ka for CH3COOH.
Ka = (H^+)(CH3COO^-)/(CH3COOH)
So (H^+) = x
(CH3COO^- = x from CH3COOH and 0.105 from CH3COOK for a total of x+0.105
(CH3COOH) = 0.170-x, now plug all of that in
Ka = 1.8E-5 (but use the value in your text) = (x)(x+0.105)/(0.170-x)
We can avoid a quadratic equation by making a couple of simplifying assumptions.
x + 0.105 = 0.105 (since x will be small)
0.170-x = 0.170 (since x will be small) so now it looks this way.
1.8E-5=(x)(0.105)/0.170
Solve for x. I get 2.04E-5 for pH = 4.69. You shouldn't take my word for anything. Confirm all of this for yourself.
But look how much easier the HH equation makes things.
The HH equation is
pH = pKa + log [(base)/(acid)]
pKa = 4.74
base = 0.105
acid = 0.170
pH = 4.74+log(0.150/0.170 =
pH = 4.68
The small difference is in rounding errors.