The Ka of hypochlorous acid (HClO) is 3.0 × 10-8 at 25°C. Calculate the pH of a 0.0385-M hypochlorous acid solution?

i got help with this problem, but i don't get where one of the numbers came from (3.4 x 10^-5)??? i tried makeing 3.0 x 10^-8 the x but i still don't get that number.

start
0.0385
change
-x. . . . .. . +x. . . .+x
at equilibrium
0.0385-x. .. x..... . .x

Ka = 3.0 x 10^-8 = [H+][ClO-]/ [HClO]= (x)(x)/ 0.0385-x

x = [H+]= 3.4 x 10^-5 M
pH = - log 3.4 x 10^-5=4.5

where is 1.155X10^-9 from?

That comes from multiplying 3X10^-8 and 0.0385 when we expand 3X10^-8(0.0385-x).

To find the pH of a 0.0385-M hypochlorous acid (HClO) solution, you can use the expression for the acid dissociation constant (Ka) and solve for the concentration of H+ ions.

The given Ka value for HClO is 3.0 × 10^-8 at 25°C.

The concentration of H+ ions can be represented as [H+]. Let's assume it to be x (mol/L). The concentration of ClO- ions will also be x (mol/L) since the ratio of H+ to ClO- is 1:1.

Using the expression for Ka, which is equal to [H+][ClO-]/[HClO], we can substitute the given values:

3.0 × 10^-8 = (x)(x)/(0.0385 - x)

Here, x represents the change in concentration of H+ ions. At the start, the concentration of HClO is 0.0385 M. During the reaction, some of it dissociates, so the concentration decreases by x, represented as (0.0385 - x) in the equation.

Next, we can solve the equation to find the value of x, which represents the concentration of H+ ions:

Multiply both sides of the equation by (0.0385 - x):

(0.0385 - x)(3.0 × 10^-8) = x^2

Now, multiply out the left side of the equation:

0.0385 × 3.0 × 10^-8 - x × 3.0 × 10^-8 = x^2

Rearranging the equation:

x^2 + x × 3.0 × 10^-8 - 0.0385 × 3.0 × 10^-8 = 0

This is a quadratic equation, which can be solved using the quadratic formula or factored. By solving this quadratic equation, we find that x = 3.4 × 10^-5 M.

Now, to find the pH of the solution, we use the formula pH = -log[H+]. Plugging in the value of [H+] = 3.4 × 10^-5 M into this formula:

pH = -log(3.4 × 10^-5) = 4.5

Therefore, the pH of the 0.0385-M hypochlorous acid solution is 4.5.

Hiya,

From what you said it seems that you understand everything up till "x = [H+]= 3.4 x 10^-5 M"
So we know that 3X10^-8 = x^2/(0.0385-x)
Bring (0.0385-x) over to the left hand side,
Hence 3X10^-8(0.0385-x) = x^2
Then we expand the left hand side,
1.155X10^-9 - 3X10^-8x = x^2
From there we bring all terms to the right hand side to give
x^2 + 3X10^-8x - 1.155X10^-9 = 0
Then we use our quadratic formula to solve for x --> x = [-b +/- sqrt(b^2-4ac)]/2a and reject the negative answer since x = concentration of H+ which cannot be negative. Then we get 3.4X10^-5 M.
Alternatively, if the math is a little too tedious, we can approximate x to be near zero since HClO is a weak acid which only dissociates very little in aqueous solution. Then we get 3X10^-8 = x^2/0.0385 which should be slightly simpler to solve.
Hope I helped!