For a solution that is 0.280 M HC3H5O2 (propanoic acid Ka=1.3*10^-5) and 0.0894 M HI, calculate the following. Concentration of H30, OH, C3H5O2-, I-

To find the concentrations of H3O+, OH-, C3H5O2-, and I- in the given solution, we will need to consider the dissociation of the acid HC3H5O2 (propanoic acid) and the dissociation of HI (hydroiodic acid). Let's break down the process step by step:

1. Dissociation of HC3H5O2:
HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)

Since the given solution is 0.280 M HC3H5O2, initially, the concentration of [HC3H5O2] = 0.280 M. When a fraction of the acid dissociates, let's assume 'x' moles, the concentration of H+ and C3H5O2- will also become 'x' moles.

So, initially, [H+] = [C3H5O2-] = 0

After dissociation, the concentration of [H+] = [C3H5O2-] = 'x'

2. Dissociation of HI:
HI (aq) ⇌ H+ (aq) + I- (aq)

Since the given solution is 0.0894 M HI, initially, the concentration of [HI] = 0.0894 M. After dissociation, the concentration of [H+] and [I-] will both become 'x' moles.

3. Expression for Ka:
The Ka value for propanoic acid (HC3H5O2) is given as 1.3 x 10^-5. The Ka expression is:
Ka = [H+][C3H5O2-] / [HC3H5O2]

Plug in the known values: Ka = x * x / (0.280 - x)

4. Calculation of x:
To solve for 'x', we need to set up an equilibrium expression and solve the quadratic equation obtained from the Ka expression.

Ka = 1.3 x 10^-5
x^2 / (0.280 - x) = 1.3 x 10^-5

Simplifying this equation, you will obtain a quadratic equation. Solve this equation using different methods, such as the quadratic formula or approximation techniques.

5. Calculate the final concentrations:
Once you determine the value of 'x', you can calculate the concentrations of [H+], [OH-], [C3H5O2-], and [I-].

[H+] = [C3H5O2-] = 'x'
[OH-] can be calculated using the equation: [H+][OH-] = 1.0 x 10^-14 (from Kw)
[I-] = 'x'

Now that you have the value of 'x', you can substitute it into the equations to find the concentrations of H3O+, OH-, C3H5O2-, and I- in the given solution.

Here is what I did for another post.

http://www.jiskha.com/display.cgi?id=1302783071