If a weak acid has a dissociation constant of 6.4 x 10^-6, what is the pH of a solution of this acid in a 0.10 molar concentration?

I still don't understand the equation you gave me. My textbook doesn't break it down like this, so I'm confused as to where to plug the 0.10 M in.

I don't remember how I worked this before. Hopefully what follows will duplicate what I did before. If you have trouble following, post the way your book does it and I'll try to explain it that way.

We'll call the weak acid HA. Then
.........HA ==> H^+ + A^-
initial..0.10M...0......0
change....-x....+x......+x
final....0.10-x...x......x

Ka = (H^+)(A^-)/(HA)
6.4E-6 = (x)(x)/(0.10-x)
Solve for x and convert to pH.

Thanks, totally still confused. I don't know how to solve for (x) since I don't know what the acid is?

The equation I'm trying to use looks like this:

1.0 x 10^-14 divided by 6.4 x 10^-6 which gives me a solution of 1.5625 x 10^-9 which can't be right!

We'll call the weak acid HA. Then

.........HA ==> H^+ + A^-
initial..0.10M...0......0
change....-x....+x......+x
final....0.10-x...x......x

Ka = (H^+)(A^-)/(HA)
6.4E-6 = (x)(x)/(0.10-x)
Solve for x and convert to pH.

You don't need to know the name of HA, everything is here.
6.4E-6 = (x)(x)/(0.1-x)
6.4E-6 = x^2/(0.1-x)
Often we can avoid a quadratic equation by making the simplifying assumption that 0.1-x = 0.1. Then
6.4E-6*0.1 = x^2 and
x = 8.0E-4 = (H^+)
pH = 3.096 which rounds to 3.10.

The inclusion of Kw in you response makes me wonder if the problem is the 0.1M SALT of a weak acid since the Kb of the salt IS Kw/Ka = 1x10^-4/6.4E-6. If your original post, however, is as above in the first post, pH = 3.1 is correct for this problem. I've gone back and re-read the post and it says nothing about the salt.

The problem was 6.4 x 10^-6. Thanks

Would that still equal a pH of 3.1?

To determine the pH of a weak acid solution with a known dissociation constant and concentration, we can use the equation for the dissociation of the weak acid and the expression for the acid dissociation constant (Ka).

1. The dissociation equation for a weak acid (HA) is:
HA ⇌ H+ + A-

2. The expression for the acid dissociation constant (Ka) is:
Ka = [H+][A-]/[HA]

To find the pH of the solution, follow these steps:

Step 1: Write out the expression for the acid dissociation constant using the given values:
Ka = [H+][A-]/[HA]
Ka = (x)(x)/(0.10 - x)

Here, x represents the concentration of H+ ions, which we assume to be small compared to the initial concentration of 0.10 M. (0.10 - x) corresponds to the remaining concentration of the unreacted HA.

Step 2: Substitute the value of Ka into the expression:
(6.4 × 10^-6) = (x)(x)/(0.10 - x)

Step 3: Solve the quadratic equation for x. Rearrange and multiply both sides by (0.10 - x):
(6.4 × 10^-6)(0.10 - x) = x^2

Step 4: Expand and simplify the equation:
(6.4 × 10^-6)(0.10 - x) = x^2
(6.4 × 10^-6)(0.10) - (6.4 × 10^-6)(x) = x^2
6.4 × 10^-7 - 6.4 × 10^-6x = x^2

Step 5: Rearrange the equation to solve for x:
x^2 + 6.4 × 10^-6x - 6.4 × 10^-7 = 0

Step 6: Solve the quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula is the most straightforward approach:
x = [-b ± √(b^2 - 4ac)] / (2a)

Using the coefficients from the quadratic equation: a = 1, b = 6.4 × 10^-6, and c = -6.4 × 10^-7, we can calculate the values of x.

Step 7: Calculate the concentration of H+ ions (x) and derive the pH:
Calculate x using the quadratic formula and select the appropriate root.
Once you have the value of x, you can calculate the pH using the equation pH = -log[H+].

It's important to note that the equation presented here assumes that x is much smaller than the initial concentration of the weak acid, 0.10 M. This simplification is valid for weak acids with relatively small dissociation constants. If x is not negligible compared to the initial concentration, the equation may need to be solved iteratively.