calculate the hydronium ion concentration in 0.125M formic acid, HCHO2

Ka= 1.8 x 10-4
can you show me the steps pls

...........HCOOH ==> H^+ + HCOO^-

initial....0.125M.....0......0
change.......-x........x......x
equil.....0.125-x......x......x

The ICE chart is set up above.
Ka = 1.8E-4 = (H^+)(HCOO^-)/(HCOOH)
Substitute from the ICE chart and solve for (H^+).

DrBob is the HCOOH---->H^++HCOO^ the HCHO2

Yes. HCOOH is formic acid. The H on the right side comes from the -COOH or an organic molecule, called a carboxylic acid, and that can ionize.

I'm stuck

Ka=1.84 x 10-4= x x/ 0.125

(1.84 x 10-4)[0.125]
2.3 x 10-5= x = [H30]
lost which way do I go

Ka = (x)(x)/(0.125-x) = 1.84E-4

We make the problem a little easier to solve if we assume x is small and that 0.125-x = 0.125. Then
1.84E-4 = x^2/0.125
1.84E-4*0.125 = x^2
2.3E-5 = x^2
x = sqrt(2.3E-5) = 0.00480M = (H3O^+).

To calculate the hydronium ion concentration in a solution of formic acid (HCHO2), you can use the equilibrium expression for the acid dissociation reaction and the Given Ka value.

The equilibrium expression for the dissociation of formic acid is:
HCHO2 ⇌ H+ + CHO2-

The Ka value represents the acid dissociation constant, which is the ratio of the concentration of the products (H+ and CHO2-) to the concentration of the reactant (HCHO2) at equilibrium.

Given:
Ka = 1.8 x 10^-4 (acid dissociation constant)
[HCHO2] = 0.125 M (concentration of formic acid)

Let's assume "x" represents the concentration of hydronium ions (H+). Since only a small fraction of the formic acid will dissociate, we can assume that the change in concentration of HCHO2 will be negligible, and thus [HCHO2] ~ 0.125 M.

The equilibrium concentrations can be expressed as:
[HCHO2] = 0.125 - x
[H+] = x
[CHO2-] = x

Now, we can substitute these values into the equilibrium expression for Ka:
Ka = [H+][CHO2-]/[HCHO2]

1.8 x 10^-4 = (x)(x)/(0.125 - x)

Since this is a quadratic equation, rearrange it to be:
x^2 = (1.8 x 10^-4)(0.125 - x)

At this point, you can solve the equation by using the quadratic formula or you can make an approximation that x is much smaller than 0.125. Since the Ka value is small, this approximation is reasonable.

Assuming x << 0.125, you can approximate the equation to:
x^2 = (1.8 x 10^-4)(0.125)

Simplifying the equation further:
x^2 = 2.25 x 10^-5

Next, to solve for x, take the square root of both sides of the equation:
x ≈ √(2.25 x 10^-5)

x ≈ 4.74 x 10^-3 M

Therefore, the hydronium ion concentration in the 0.125 M formic acid solution is approximately 4.74 x 10^-3 M.