Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.
a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5
Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?

Question

Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.

a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5

Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?

Accepted Answer

All right. Hello everyone. So this question says that the K A one and K A two of a 0.20 molar weak Polypro acid H two A are 1.5 multiplied by 10th, the negative 4th and 2.0 multiplied by 10th, the negative fifth respectively. Part one says to determine the amount of H 30 positive contributed by the first ionization step. Part two says to determine the amount of HDO positive contributed by the second ionization step. And part three is asking to determine when the contribution from the second step can be ignored. OK. So first, let's go ahead and start by answering part one, which is the dissociation or rather the concentration of HCO positive from the first ionization step. The reaction for that is as follows, we have our weak Polypro acid that's H two A combining with water or dissolving in water to produce hydro num ions H 30 positive as well as the conjugate base of H two A which would be H A negative. This is our first ionization. And from here we can go ahead and calculate or construct, I should say our ice table recall that ice tables are useful in understanding the changes in concentrations during an equilibrium process. So I stands for initial C stands for change and E stands for equilibrium. Now do keep in mind that the change in the concentration of water is not going to be trapped because water being a liquid is not going to be included in the expression for the equilibrium constant. But otherwise the other species in this reaction are aqueous and therefore will be considered in the equilibrium concept. So from here in the column representing the initial values H two A has a concentration of 0.20 molar, I join him. The hydro num ion and H A negative are initially going to have a concentration of zero molar because they are the products. And now as for the change, the concentrations of H three positive and H A negative are both going to increase by X. On the other hand, the initial concentration of H two A that's 0.20 is going to decrease by X. So the equilibrium row which represents the changes in their concentrations after the equilibrium process is essentially going to be the sum of both the initial and the change columns. So for H two A, for example, the equilibrium concentration would be 0.20 subtracted by X. Whereas for H 30 positive and H A negative, they're going to be simply equal to X. So now let's go ahead and recall the expression for the first equilibrium constant. That's K A one recall that an equilibrium constant is going to be equal to the concentrations of products divided by the concentrations of reactants excluding any solids and liquids. So here K A one is equal to the concentrations of both H 30 positive and H A negative modified or rather divided by the concentration of H two A substituting each variable here with our equilibrium concentrations determined from the ice table. This expression is equal to X squared divided by 0.20 subtracted by X. So here in order to try and make our calculations a little bit easier, I want to go ahead and check if X is negligible in the denominator. And I would do that by calculating the concentration of H two A divided by K A one. So that's 0.20 divided by 1.5 multiplied by 10 to the fourth power, excuse me, negative fourth power. And that equals about 1333. And so because this number here is greater than 500 X in the expression of the denominator is going to be negligible. So my new expression can be approximated as X squared divided by 0.20 for K A one. So now we can go ahead and use this to solve K A one is equal to 1.5 multiplied by 10 to the negative fourth power and this equals X squared divided by 0.20. So now to go ahead and isolate X, I'm going to multiply both sides of my expression by 0.20. And then to go ahead and solve for X, I'm going to take the square root of that product. So overall X is equal to the square root of 1.5 multiplied by 10 to the negative fourth multiplied by 0.20. And this ultimately equals 5.4772 multiplied by 10 to the negative third moller. This is the amount of H 30 positive contributed by the first ionization set. So after rounding two two significant figures, the answer to part one is going to be 5.5 multiplied by 10 to the negative third molar. However, for the, for the sake of the second step, I'm going to go ahead and use the ungrounded number because the ungrounded value represents the amounts of both HCO positive and H A negative for the second ice table because now we have to recall the second dissociation equation. So here I'm just going to go ahead and write as a reminder that X is equal to the concentration of H A negative, which is also equal to the concentration of HCO positive. So now let's go ahead and talk about our second dissociation or a second ionization step in which now our acid is going to be H a negative which dissociates in water to produce H 30 positive and A two negative. So using a new ice table, the concentration or rather the value for X that we calculated in the first step represents the initial concentrations of both H negative and H 30 positive. Once again, water is excluded from the equilibrium constant. So we're not going to take into consideration its change in concentration. So for initial, the initial concentration of H A negative is zero point 005477 and the same is true for the initial concentration of HCO positive, the initial concentration of A two negative is currently going to be zero. So now to represent the change in these values, once again, the two products are going to increase in concentration. But now we're going to let Y represent the change during the second ionization step. This means that the concentration of H A negative which is a reactant is going to decrease by Y. Therefore, the equilibrium concentrations are going to be essentially combining the first two rows here. So the equilibrium concentration of H eight negative is going to be 0.8 005477 subtracted by Y. And for HCO positive, its going to be 0.005477. Add it to Why? And for a two negative its simply going to be Y. So from here we can go ahead and make substitutions to the expression for K A two, K A two is equal to the concentrations of both HCO positive and A two negative divided by the concentration of H A negative. So from here, we can go ahead and substitute the values for K A two and the equilibrium concentrations of all of our species here. So that's 2.0 multiplied by 10 to the negative fifth power equal to Why multiplied by 0.005477 added to Y divided by 0.005477 subtracted by Y. So in this particular case to go ahead and simplify this expression, I'm going to multiply both sides of the expression by 0.005477 subtracted by Y. This means that I would have to distribute the value of K A two into both of those terms. And in the numerator I can distribute or Y into both terms of the equilibrium concentration of H 30 positive. So from here, we can simplify this expression as such. We have 1.0954 multiplied by 10 to the negative seventh power subtracted by two, multiplied by 10 to the negative fifth, multiplied by Y equals 0.005477 multiplied by Y added to Y squared. So from here, I'm going to go ahead and bring all terms into one side of the expression. So that one side of the equation is equal to zero. And so after rearranging my expression and combining like terms as needed, we get Y squared added to 5.497 multiplied by 10 to the negative third power multiplied by Y subtracted by 1.0954 multiplied by 10 to the negative seventh equals zero. From here using the quadratic formula, we can go ahead and solve four values of why. After doing so Y is equal to 1.985 multiplied by 10 to the negative fifth power. The quadratic formula does give us an extra solution. That's negative 5.51685 multiplied by 10 to the negative third. However, a concentration cannot be negative. So we're going to keep our positive solution. So here we have the value for why. But it is worth mentioning that the concentration of H three plus generated by the second ionization step is not just going to be equal to Y rather, it's going to be equal to 0.005477 added to the value of Y that we just calculated which is 1.985 multiplied by 10 to the negative fifth power. And this is going to ultimately equal 0.005497. So now let's go ahead and answer part three which is attempting to see at what point we can ignore the cons the contribution of the second step. So here we have to consider the percent contribution of that second ionization that's going to be equal to Y divided by X multiplied by 100%. When substituting the values for both Y and X and multiplying this by 100% this is going to equal a percent contribution of the second ionization equal to only 0.36% which is negligible. What this means is that it's not going to affect the ph calculation for the solution. And this is confirmed after considering the fact that calculating the ph using the hydro num ion concentration from both steps gives you approximately equivalent values. And so now weve arrived at our final answer. So for parts one and two, the concentrations of hydro num ion after the 1st and 2nd ionization steps are equal to 5.5 multiplied by 10th, the negative third molar after rounding to two significant figures. As for the third part of this answer, the contribution from the second step can be ignored when calculating the ph of the solution. And so with that being said, thank you so very much for watching. And I hope you found this helpful.

Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.
a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5
Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?

Verified Solution

Key Concepts

Polyprotic Acids

Dissociation Constants (Ka)

Hydronium Ion Concentration ([H3O+])

Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given here.a. Ka1 = 1.0 × 10–4; Ka2 = 5.0 × 10–5Calculate the contributions to [H3O+] from each ionization step. At what point can the contribution of the second step be neglected?

Verified Solution

Key Concepts

Polyprotic Acids

Dissociation Constants (Ka)

Hydronium Ion Concentration ([H3O+])

Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.
a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5
Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?