It asks us what is the original molarity of a solution of weak acid with a \( K_a \) of \( 4.7 \times 10^{-3} \) and a pH of 4.12 at 25 degrees Celsius. Now, the key here is realizing that we're dealing with a weak acid. Remember, we know it's a weak acid because its \( K_a \) value is less than 1. With a weak acid, we're going to say that our generic weak acid, we can say represents weak acids, weak bases react with water in order to establish the equilibrium. Remember, the acid here according to the Bronsted-Lowry definition will donate an \( H^+ \).
It's going to become \( A^- + H_3O^+ \). We have initial change equilibrium. Water is a liquid. Remember, liquids and solids are not included within an ICE chart. Here, we're told to find the original molarity, so that's our formal concentration.
We don't know what it is, so we'll just place \( f \) here. These initially are 0. Remember, we're losing our reactant over time in order to produce products. This will be minus \( x \) plus \( x \) plus \( x \). Bring down everything, \( f - x, x \).
Knowing this, we set up the equilibrium expression as \( K_a = \frac{A^- \times H_3O^+}{\text{Reactant}} \). So it's \( K_a = \frac{A^- \times H_3O^+}{\text{Reactant}} \). Again, we're ignoring water because it's a liquid. Our \( K \) is told to be \( 4.7 \times 10^{-3} \). At equilibrium, both of my products are equal to \( x \). So if they're multiplying each other, that's going to be \( x^2 \) on top divided by \( f - x \).
What we have to do here is to figure out that formal concentration. They tell us the pH of the solution. Remember, if we know pH, then we know \( [H_3O^+] \) concentration. With that \( x \) variable, you can plug it into here and plug it into here. We'll have all the variables we need and the only missing one will be our formal concentration which we'll have to solve for.
Alright, so we're going to plug in \( 10^{-4.12} \). When we do that, that gives me an answer of \( 7.59 \times 10^{-5} \). So again, this is equal to our \( x \) variable. So what we're going to have here is \( 4.7 \times 10^{-3} = \frac{(7.59 \times 10^{-5})^2}{f - 7.59 \times 10^{-5}} \).
So all we need to do again is figure out our formal concentration. So cross multiply these two. So this will distribute into here and distribute into here. So when we do that, that's going to give me at this point \( 4.7 \times 10^{-3} \times (f - 3.56531 \times 10^{-7}) = 5.7544 \times 10^{-9} \).
Remember, we need to isolate our formal concentration. So we're going to add \( 3.5631 \times 10^{-7} \) on both sides. So this cancels out. Bring down \( 4.7 \times 10^{-3}f = 3.62286 \times 10^{-7} \). Isolate your formal concentration, so divide both sides by \( 4.7 \times 10^{-3} \).
So here at the end, our formal concentration equals \( 7.71 \times 10^{-5} \) molar. That there represents our original or formal concentration of our weak acid. Remember, things to look out for. If we have a weak acid, we'll have a \( K_a \) less than 1. If we have a weak base, we'll have a \( K_b \) less than 1. Here, if they're telling us pH, that can give us \( H_3O^+ \). If they gave us pOH, then that will give us \( OH^- \). These are key things that you need in order to figure out whether they're asking for the concentration of the original weak acid or one of its products.
