Problem

What two common assumptions can simplify calculat…

01:15

Problem 49 Hard Difficulty

What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?

Answer

1) $c_{0}\left(H^{+}\right)=0$
2) $\Delta c(H A)=0$

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Chemistry 102

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Video Transcript

problem 49 from Chapter 14 is asking which assumptions can be made to simplify the calculation of equilibrium concentrations in a solution of a weak ass. And so here I'm just drawn a, um example, um, for us to examine and understand why we can make these assumptions. So this, um, equation involves nitrous acid reacting with water to give hydro Nia my Hans and nitrogen oxygen. So, um, let's just start with the initial concentration of our weak acid, which we will be just arbitrarily choose. Go 0.5. And this is all in Moeller moles. The initial this water is the, um, solvent. So we just ignore that, um And then this is where our first assumption comes in is that the contribution of the solvent of water to the initial concentration of hydro knee, um ions is negligible. So we can write approximately zero here, and this is our product. So our initial concentration is zero. And for the change, it's going to be minus X. And here we're going to have plus X and plus Dex and then for a concentration at equilibrium. This is going to be a final concentration will be no 0.5 minus X and for, um Hi, Joanie. Um, will have x on. And for nitrous oxide, we will also have X. So when we are solving for our organization constant work, eh? Um, we include the, um We divide the product, the concentration of the product by the concentration of the reactant. So that looks like this Okay is equal to concentration of our product, which is the hydro Nia, my on and nitrous oxide in our example, divided by be concentration of nitrous acid. So if we go back and refer to our ice table that we created on the previous page how we're, um numerator is going to be x times X, and our denominator is going to be 0.5 minus x so we can go ahead and put that in X. I'm six invited by. So your 0.5 Mueller minus X And here is when the second assumption comes in, it's message X hears of this. Second assumption comes in. So our initial concentration of acid is group is quite high. It 0.5 bowler, and, um, we can assume that I were equilibrium concentration and our X is going to be much, um, much lower than our initial concentration. So we can actually get rid of this X here, making our equation much simpler. Um, so then it's just X squared, for example, X squared, divided by 0.5. And again, you can make this assumption because you're assuming that the equilibrium, concentration or the change constant the X value is going to be much, much lower than your initial concentration. And the way you verify that is once you solve for X using the given K A. You can go ahead and plug your ex value in and see if it is, um, 5% or less than the initial concentration. So this assumption only works if your ex value is less than 5% of your initial concentration. If it's more than five person, then you have to move on and use the quadratic equation. You can't man a negate this X over here so that two assumptions are that water does not contribute to a, um, increase in the hydro ni am I on and that the change or the ex constant value of X, is going to be much less than your initial concentration

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