Calculate the [H3O+] and pH of each H2SO4 solution. At approximately

what concentration does the x is small approximation

break down?

a. 0.50 M b. 0.10 M c. 0.050 M

Video transcript

Hello everyone. Today, we have the following problem. Calculate the concentration of hydro ions and the ph of a strong Dipro acid with the falling PK value at the given concentrations at which given concentration or concentrations does the assumption exit negligible, not applicable. So we have one being one molar, two being 0.10 molar and three being 0.01 molar. So since the acid is not specified, we can just use this following acid as a generic Dipro acid. And so since the first association reaction is full as you're not given that the initial concentrations will be based on the products of the first reaction. So when we have our first reaction of our Dipro acid, this will be aqueous, we will place it in liquid water. This will be a reversible reaction and this will dissociate into an aqueous proton and the following conjugate base which is aqueous in nature, then the second reaction will take that conjugate base that was formed and it will also still be in water and a reversible reaction to dissociate into an aqueous proton and its final conjugate base form which is aqueous. So now we have to construct an ice chart and our equilibrium equation. So for equilibrium equation, we take our acid ionization constant and we set that equal to the concentration of our products in the numerator. We divide that by the concentration of our reactants in the denominator. Now we were, this would then add to be the causation of products which would be our acid as for the concentration of our acid multiplying. And we are, we are using the second equation. So be the acid which has a negative two charge multiplied by the concentration of hydro lium ions. And then for our reactants, we have our conjugate base. So we were given the PK value of 2.5 times 10 to the negative two. We will hold on to that setting up our ice chart. We have our equation here. So we have our, let's actually rewrite this. So we have our conjugate base and aqueous reacting with water, liquid water and a reversible reaction to form aqueous proton and that conjugate base. So our ice tables, we have our initial in the first for the first row, initial concentrations followed by the change in concentrations and in the equilibrium values. So our initial concentrations of this first reactant is one molar. And this is going based off the example that we have for one since water is a liquid, we will not be including it or since water is in its liquid form, we will not be including this. And then our concentrations of our products for the initial concentrations are zero for the protons or it's zero for the conjugate base. But it's one or the hydro num. Of course, the change is going to be from reactants to products. So we can represent that for the reactants by minus X and then plus X for both products. Now, equilibrium is essentially row I plus row C. So we have one subtracted by X fire reactants. And then we have X, we have X four hour conjugate base and then we just have X or hour or one plus X for our hydro num. Now, for the individual concentrations, we will start with one molar. What we need to do is you need to set this PK a value that we were giving up. So 10, 2.5 times 10, 782 we will make that equal to the concentration of our products, which will be the X multiply body, one plus X divide that by the concentration of our reactants, which is one minus X. If we were to fill in our values or if we were to solve, we would get the following where the X would distribute into the one and the following X. And then we will cross multiply to get 2.5 times 10 to the negative two multiplied by one and then multiplied by negative X, which would essentially give us. So that would be equal to. And then we would have X plus X squared, which we can subtract 2.5 times 10, the negative two from both sides to give us the following. Now, this has to be solved using the quadratic equation. So if we recall that that was going to be our negative B multiplied by this, it's going to be a negative B plus or minus the square root of negative B squared or just B squared subtracted by four AC. And then we divide that by two A. So for our first one molar concentration, we have our negative B which is negative X. So that would just be, that should be plus. So we would have negative one followed by E plus and minus square root of B squared. So we have just one squared subtracted by four multiplied by A which is one in this case. And then we have C which can be represented, there should be 10 to the negative two. So we have 2.5 and this is negative times 10, reaching the power of negative two, giving us values of 0.024 and neg negative 1.0. But we will use the positive value. And since the positive value is 0.024 it X is negligible in this case. So X is negligible in the first concentration, we will essentially repeat the process for the next two concentrations. So we have 0.10 molar for this one, we will set up the exact same equation where in we have our PK value is equal to, we will have our X multiplied by 0.10 plus X divided by 0.10 minus X. And we solve a four, our equation, we will arrive at another quadratic equation such that X squared plus 0.13 X subtracted by 2.5 times 10 to the negative three will be our value. The quadratic equation if we set it up will be negative 0.13 plus or minus 0.13 squared subtracted by or multiply A which is one and then multiply it by our C which is negative 2.5 times 10 to the negative three. This will all be over two A which would just be two times one or two. And this actually for the first equation for one molar that should be divided by two as well. So for our 0.10 molar, we get a, we get concentrations of 0.17 a 0.0 17 molar and then we get negative 0.15 molar. However, as before we use the positive value and this would round up to be 0.02. If we use significant figures, two significant figures and X will not be negligible. In this case, it should be 12 and that value is too small and we can say X is not negligible. So X is not negligible for part two or part three, we will have once again, the same equation of 2.5 times 10 to the negative two. This time it will be equal to X times the concentration of 0.01 zero divided by. And this would be plus X as well divided by 0.010 minus X which will simplify to X squared plus 0.035 X minus 2.5 times 10 to the negative fourth plugging that into our quadratic equation, we would get negative B which is 0.035 plus or we actually do this below. So we'll have our negative 0.035 plus or minus the square root of 0.035 squared subtracted by four multiplied by A which is one and then C which is negative 2.5 times 10 to the negative fourth. Of course, Oliver two A which is two times one or just two. Solving for X, we get a, a number of X is equal to 0.0061 and negative 0.041. However, we want to use the positive value this value here. If we were to add to the initial concentration of one, we would get 1.024 molar. And then we can conclude that X is negligible for our third part. So X is only not, so, so the X is negligible is only not applicable to the second part or the concentration of 0.10 molar. And without we have solved the problem overall, I hope it's helped. And until next time.