Pyridinium bromide (C5 H5 NHBr) is a strong electrolyte that dissociates completely into C5 H5 N

step 1 Chapter 16, Problem 82 from Chemistry, the Central Science. The question is asking if we have pr

Pyridinium bromide (C5 H5 NHBr) is a strong electrolyte that...

Pyridinium bromide $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)$ is a strong electrolyte that dissociates completely into $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$ and $\mathrm{Br}^{-} .$ An aqueous solution of pyridinium bromide has a pH of $2.95 .$ (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the $K_{a}$ for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?

Pyridinium bromide $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)$ is a strong electrolyte that dissociates completely into $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$ and $\mathrm{Br}^{-} .$ An aqueous solution of pyridinium bromide has a pH of $2.95 .$
(a) Write out the reaction that leads to this acidic pH.
(b) Using Appendix D, calculate the $K_{a}$ for pyridinium bromide.
(c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium
cation atequilibrium, in units of molarity?

Key Concepts

Equilibrium Calculations in Acid-Base Systems

This involves setting up and solving equilibrium expressions based on the acid dissociation constant (Ka) and the initial conditions determined by complete dissociation. It is a fundamental technique for quantifying the concentrations of all species present at equilibrium in an acid-base reaction.

Acid Dissociation Constant (Ka)

The Ka value quantifies the extent to which an acid releases protons in solution. It is essential for determining how much of the conjugate acid dissociates into its ions and for calculating equilibrium concentrations in acid-base systems.

pH and Hydronium Ion Concentration

pH is defined as the negative logarithm of the hydronium ion concentration in a solution. Understanding this relationship allows one to translate pH measurements into actual concentrations of hydrogen ions, which is critical in evaluating acid-base equilibria.

Strong Electrolyte Dissociation

This concept refers to substances that, when dissolved in water, break up completely into their component ions. In acid-base reactions, knowing that a compound dissociates fully helps establish the initial concentrations of the ions that will participate in further equilibrium processes.

Conjugate Acid Hydrolysis

When a weak base's conjugate acid is present in solution, it can react with water to donate a proton, forming hydronium ions. This process, known as hydrolysis, explains how a salt derived from a weak base can make the solution acidic.

Transcript

00:01 Chapter 16, problem 82 from chemistry, the central science.

00:06 The question is asking if we have pranineineineine on bromine is a strong electrolyte that disassociates completely into c5h5nh plus and br minus.

00:22 An aqua solution of bromine has a ph of 2 .95.

00:27 So they're asking us to first write out an equation.

00:31 And that would give us an acidity, ph acidity.

00:36 So what we're going to do is we're first going to write the dissociation of the molecule.

00:44 So we have c5h5nhb r, and whenever you dissociate that, you're going to get your c5h5nh plus, plus br minus.

01:01 And the reason why i wrote this out first is because we're going to, me that a little bit later.

01:05 And so now we're going to figure out which way we're going to get an acidic ph.

01:09 And so for something for you to have something acidic, you have to have an h plus in your final item.

01:15 So we want one of these two products on this side to be able to accept a hydrogen.

01:20 And the one that's going to accept the height, i mean, except an oh.

01:24 And the one that's going to accept an oh is our cshs in h plus.

01:31 So the reason how we know that is because it is a positive number and oh h is a minus number and when you add them together they're going to give you um that take that oh h and then leave you with an h plus so our equation will be c5 h5 n h plus plus h2 o gives you c5 h5 n h o h plus because if we would have done br, we would have br plus h2o gives us hbr plus oh minus.

02:26 And we were running an h plus, not an oh, because we want something acidic.

02:32 So now since we wrote the equation that leads to the city of the ph, we now can be able to use that in order to help calculate our final equilibrium constant in part c.

02:44 So b is now asking us to calculate the ka for the equation for a printer -blown -mind, and so to do that, we need to know that k -a and k -b is equal to our k -w.

02:59 Our k -b, whenever you're looking up in the back of the book, is equal to 1 .7 times 10 to the negative ninth.

03:17 And so now we're just going to solve for k -a.

03:23 So k -a is equal to our k -w divided by k -v.

03:30 So 1 times 10 to negative -14, over 1 .7 times 10 to negative 9th.

03:38 And when you plug that into your calculator, we get 0.

03:52 We get 5 .8 times 10 to negative 6.

04:10 And that is what our ka is equal to.

04:12 So that is our second part of our answer.

04:17 Now, part c was asking us to find a solution of the permitting grime that has a ph of 2 .95.

04:28 And what is the concentration of the cadion at equilibrium? so to be able to do that, we just rewrite that equation that we did in the first part.

04:37 So we have our c5h5nh plus plus h2o goes to c5h5 in h0 plus.

05:04 And now we just write an ice table.

05:10 And we're wanting to find the concentration of the caddite at equilibrium in units of molarity.

05:19 So to be able to do that, we then are going to do m if we're here because we don't know our initial concentration.

05:31 And then also water does not play a role in our equation and then we started off with zero and zero here...

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