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Ch.16 - Acid-Base Equilibria

Chapter 16, Problem 18a

Identify the Brønsted–Lowry acid and the Brønsted– Lowry base on the left side of each equation, and also identify the conjugate acid and conjugate base of each on the right side. (a) HBrO1aq2 + H2O1l2ΔH3O+1aq2 + BrO-1aq2

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Hello everyone today, we are being asked to consider the following reaction hyper cleric acid plus water gives us hydro knee. Um And chlorate ions determine the bronze said Laurie acid and bronze said Laurie base on the left side of the equation and then determine their corresponding conjugate acid and conjugate base on the right side of the equation. So first we're going to know that either this hipaa chlorate or this water will be the bronze Hillary acid or base and then the hydro ni um or chlorate will be the conjugate base or conjugate acid. So from our definitions we know that a bronze stead Laurie acid abbreviated by B. L. A. Is going to be a proton donor. So it's going to donate a proton to a chemical species. And so if we look closely at this question on the left, with these two options, we can see that this H C L 04 on the left has a proton but on the right it is missing one from that species. So therefore that is the the proton or brown said Laurie acid. For our bronze celery base this is going to be our proton except er So it's going to accept a proton from a chemical species. If we look at our chemical equation we can see that water on the left here gained a proton on the front that's shown on the right. Making water. Our brown said Glory base are conjugate acid will just be the reversal. So for a conjugate acid it's going to be one hydrogen greater than our water giving us our hydro ni um and then for our conjugate base, this is going to be a hydrogen less than our original bronze ted Laurie acid, Making that chlorine cl of 4 -. And these are our final answers. I hope this helped, and until next time.