Ammonia 1NH32 has base dissociation constant 1Kb2 of 1.8 * 10-5. What is the concentration of an aqueous ammonia solution that has a pH of 11.68? (LO 16.11) (a) 0.28 M (b) 3.6 M (c) 9.0 * 10-3 M (d) 1.3 M

Video transcript

Welcome back, everyone. We need to calculate the concentration of ethyl ammonium bromide in a ethyl amine, methyl ammonium bromide solution. If the concentration of ethyl amine is 0.14 Mueller and the ph of the solution is 10. Note that the KB four ethyl amine or it's based association constant is the value 4.47 times 10 to the negative fourth power. Let's begin by recalling that ethyl amine C two H five and H two is a weak base and therefore, it only partially dissociates. And so in water, it's going to partially dissociate in an equilibrium where we form our ethel ammonium Kati on C two H five N H three plus, since it will gain a proton from water acting as an acid, and we will also have as a result. Our second product being hydroxide. And so our ethel ammonium Kati on is our conjugate acid and hydroxide is our conjugate base for water. So now with this reaction outlined, notice that our salt mentioned in the prompt methyl ammonium bromide C two H five N H three BR will fully dissociate and dissolve in water into its constituent ions So we'll have C two H five and H three plus, which again is our conjugate acid of our ethyl amine weak base. And we will also have our bromide an ion form as a second product. So recall that with any salt, it's always going to fully dissociate into its ionic components. And so if we are able to understand this reaction, then this tells us that our concentration of our salt, methyl ammonium bromide should equal our concentration of our bromide an ion and our concentration of our ethyl ammonium catalon. And that's because our salt fully dissociates into these products. And so we have a 1 to 1 ratio between our salt and its ion products allowing for their concentrations to all be equal. So in order to get to our final answer, we're going to need to find our concentration of our methyl ammonium catalon. And if we recall our KB expression, which is equal to our concentration of our products. So, in this case, methyl ammonium C two H five and H three plus in the numerator multiplied by the concentration of our second product hydroxide divided by our concentration of reactant where we don't include our liquid water, but we will include our ethyl amine C two H five N H two in the denominator. So let's first find our concentration of hydroxide Recall that were given a ph equal to 10 for our solution. And recall that we can find the concentration of hydro ni um by taking 10 to the negative Ph. And so, in this case, we would have 10 to the negative 10th power giving us our concentration of hydro ni um equal to one times 10 to the negative 10th power moller. Next, let's find our concentration of hydroxide, which is what we need. And this is going to equal our ion product of water K W divided by our concentration of hydro ni. Um Since we know our concentration of hydro knee, um and we should recall that in our numerator are ion product of water is one times 10 to the negative 14th power molars squared. From our notes in our denominator, we can plug in our solved concentration of hydro knee um one times 10 to the negative 10th power moller. Now notice that in our numerator are KW for water is Mueller squared. We know that it's Mueller squared because recall that our ion product of water K W is equal to our concentration of hydro ni um in moller times our concentration of hydroxide and moller. And so that's how we get square moller units for our numerator KW. So now in our calculators plugging in this quotient, we're going to come up with a value for our concentration of hydroxide equal to one times 10 to the negative fourth power moller. So now we're going to utilize this concentration of hydroxide in our KB formula. So going to our KB formula were given the based association constant for ethylene in as 4.47 times 10 to the negative fourth power equal to recall. Again, R K B expression is our concentration of products. In this case, our concentration of ethylene we need to solve for or sorry, our concentration of methyl ammonium. So that's our numerator, C two H five and H three plus, we can plug in our concentration of hydroxide which we just solve two B one times 10 to the negative fourth power moller. And in our denominator, we have our concentration of our reactant mean Going back to the prompt were given that concentration as 0.014 moller for ethyl amine in our denominator. And now with this quotient plugged in, we want to simplify as much as we can. So we need to isolate for our concentration of our numerator ethel ammonium. So we're going to multiply both sides by our denominator, 0.014 Mueller So that it canceled out on the right hand side and we're going to divide both sides by one times 10 to the negative fourth power moller so that it can cancel it on the right. And what this will simplify too is R K B 4.47 times 10 to the negative fourth power multiplied by 0.14 moller divided by One times 10 to the negative 4th power moller. This is going to be set equal to our concentration of ammonium C two H five and H three plus. So, in our calculators, we're gonna plug in our quotient on the left and we'll find that our concentration of methyl ammonium on the right is equal to a value of 0.6258 moller. And since above, we understood that our concentration of our ethel ammonium bromide salts is equal to our concentration of our ethel ammonium catalon. And our bromide an ion, we can say that this is also equal to our concentration of our ethel ammonium bromide salts. So C two H five and H three BR Also has a concentration of 0.06258 Moller. And we can actually round this To about two sig figs as 0.063 molar for our concentration of our salt. And so for our concentration of ammonium bromide, we've determined our final answer to be 0.063 moller corresponding to choice c in the multiple choice as are correct final answer. So I hope that this made sense and let us know if you have any questions.