Calculate the [H₃O⁺] and pH of each polyprotic acid solution.
a. 0.125 M H₂CO₃

Question

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. a. 0.350 M H₃PO₄

Accepted Answer

Hey everyone, we're asked, what is the ph and concentration of our protons of a 0.24 Mueller polyp protic acid First looking at R. K. S, we're going to use our first K. A. Since it is much larger than our 2nd and 3rd K. Now let's go ahead and write out our reaction. So we have our polyp protic acid which will symbolize as H3 a. And we react this with water. We then form the conjugate base of our product acid, which will be H two a minus plus our hydro ni um ions. We can then go ahead and create our ice chart. Initially, we were told that we had 0.24 Mueller of our polyp protic acid and we had zero of our products. Initially we can go ahead and disregard our water since it is a liquid and our change is going to be a minus X. In our reacting side. Since we're losing reactant and a plus X in our product side. Since we're gaining products At Equilibrium, we're going to have 0.24 Moeller -X. For our react inside an X. And an X in our product side. For our K. A. one. We know that this is going to be our products over our reactant and this is going to be equivalent to our K. A. One of 6.3 times 10 to the negative third First. Let's go ahead and check if we can disregard our X. And we can do so by taking our 0.24 And dividing it by our K. A. one of 6.3 times 10 to the negative third. If we get a value Less than 500 then we cannot disregard it. And in this case we got a value of 37 Which is in fact less than 500. So we have to keep our X solving for R. X. We can further simplify this and we end up with X squared over 0.24 minus X equals 6. times 10 to the negative third. We can go ahead and multiply both sides by 0.24 -1. We then get X squared equals 6.3 times 10 to the negative third Times 0.24 -X. And since we do need to use our quadratic equation we can rewrite this as X squared plus 6.3 times 10 to the negative third. X minus 6.3 times 10 to the negative 3rd times 0.24. Now that we have it in this form, let's go ahead and use our quadratic equation to solve for X. Using our quadratic equation, we end up with X as 0.03586. So this means our concentration of hydro ni um ions is 0.3586, Mohler. Now in order to find our ph we can then take the negative log of our concentration of hydro knee um ions which is 0.3586, And this will get us to a ph of one 0.45. So our final answers here are going to be a ph of 1. and a concentration of hydro ni um ions of 0.36 Mohler, since we have two significant figures, so I hope this made sense and let us know if you have any questions.

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. a. 0.350 M H₃PO₄

Verified Solution

Key Concepts

Polyprotic Acids

Dissociation Constants (Ka)

pH Calculation

Calculate the [H3O+] and pH of each polyprotic acid solution. a. 0.350 M H3PO4

Verified Solution

Key Concepts

Polyprotic Acids

Dissociation Constants (Ka)

pH Calculation

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. a. 0.350 M H₃PO₄