Calculate the [H₃O⁺] and pH of each polyprotic acid solution.

b. 0.125 M H₃C₆H₅O₇

Video transcript

Welcome back everyone. What is the Ph and the concentration of hydro num for the poly protic acid solution provided below. We're given 0.15 molar C seven H 603 with two K A values. Meaning it is a Dipro acid. Essentially, what we want to understand is that assets, if they are weak assets and we are given their low K values, they will in water to produce hydro num. So if this is a dip protic acid, we can just use a general form of H two A, right? We understand that there are two acidic hydrogens within this acid and A is just a conjugate base. So essentially what we're going to do is just consider the first ionization step where we are losing the first hydrogen to what are to produce hydro, we are producing H A minus and hydro. And we essentially have to recall that whenever we want to identify the concentration of hydro and ph, we are essentially constructing an IC table in terms of molarity here. So initially, we have a concentration of 0.15 molar and we can just say that we have zero, approximately zero for hydro, the change would be negative X and positive X for the product side. And at equilibrium that would be 0.15 minus XX and X. And we know that this is the first ionization step. So K A one would be equal to X square divided by 0.15 minus X. We're simply taking the products multiplying their concentrations and dividing by the concentration of the reactant equilibrium. And our essential, we want to see whether or not we have to solve the quadratic equation to do that. We are going to take the initial concentration and divide it by K A one. The initial concentration is 0.15. And we're going to divide it by 8.3 multiplied by its sense. The power of negative fifth. What we notice is that this is much, much greater than 1000 meaning negative X in the denominator is negligible. And if we solve this equation, we can essentially simplify it as ke one is approximately equal to X squared divided by 0.15. So our X is simply square root off the product between the initial concentration and K A one which is the concentration of hydro at equilibrium. We're going to neglect the second ionization step, right? Because as we understand this is a very weak acid and we can also neglect the value of X in the denominator. The. Therefore, the second ionization step will not matter because K A two is also much lower than K A one. So if we solve for X, we get a value of, we need to use two significant figures. And that'd be 3.5 multiplied by sense, the power of negative third molar. Now this would be the concentration of hydro and now our ph is negative log of the concentration of hydro by definition. So we're just going to take the previously calculated value in order to avoid the propagation of air. We're just going to substitute square root of the product between the concentration and K A one. And this gives us the P value of 2.45. So we have our concentration of hydro and also the ph of the solution which is 2.45. So let's label our answers. That would be it. Thank you for watching.