Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. a. 0.15 M HF

Video transcript

Hey everyone, we're asked, what is the ph of 0.450 molar of Hydrofluoric solution. Can we apply the assumption the X is small 1st. Let's go ahead and write out our reaction. So we have Hydrofluoric acid plus our water. And this is going to form our hydro ni um ions plus our flooring an ion. We can go ahead and create our ice chart. And initially we were told that we had 0.450 molar of Hydrofluoric acid and we can go ahead and disregard our water since water is a liquid. And for our products, we started off with zero of our products, initially our change is going to be a minus X for our Hydrofluoric acid. Since we're losing reactant and a plus X for our products, since we're gaining products, our equilibrium is going to be 0.450 minus X. For Hydrofluoric acid. And for our products we're going to have X. And an X. Next we can go ahead and solve for X. So if we look at the back of our textbooks, RKA for Hydrofluoric acid is going to be 6.6 times to the -4. And this is going to be equal to our products which is X&X over our reactant, which is 0.450 -1. Now, in order to disregard our X. And our denominator. We can go ahead and check if it is small enough to do so we're going to take our 0.450. And we're going to divide this by R. K. A. Of 6.6 times 10 to the negative four. And we end up with a value of 682. And if our value is greater than 500 then we can safely disregard our X. and assume that it is small. So in this case our X. Is small enough solving for X. We can go ahead and simplify this a bit further. So we get 6.6 times 10 to negative four equals X squared over 0.450, Multiplying both sides by 0.450. We end up with X squared equals 2.97 times 10 to the -4. Taking the square root of both sides. We end up with a value of x equals 0.01723. And this is going to be our value for our hydro knee um ions and our flooring and ion solving for R. P. H. We can then take the negative log of our concentration of hydro ni um ions. So this will be negative log of 0.01723. And we end up with the value of 1.76 as our pP so this is going to be our final answer. And also we were asked if X was small enough to simply disregard it. So yes it was small enough. Now I hope that made sense And let us know if you have any questions