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If $2.50 \mathrm{g}$ of $\mathrm{CuSO}_{4}$ are dissolved in $9.0 \times 10^{2} \mathrm{mL}$, of $0.30 \mathrm{M} \mathrm{NH}_{3},$ what are the concentrations of $\mathrm{Cu}^{2+}$ $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+},$ and $\mathrm{NH}_{3}$ at equilibrium?

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$$\begin{array}{l}{\left[N H_{3}\right]=0.2304 M} \\{\left[C u\left(N H_{3}\right)_{4}^{2+}\right]=0.0174 M} \\{\left[C u^{2+}\right]=1.235 \cdot 10^{-13} \mathrm{M}}\end{array}$$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

Rice University

University of Toronto

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

24:14

In chemistry, a buffer is a solution that resists changes in pH. Buffers are used to maintain a stable pH in a solution. Buffers are solutions of a weak acid and its conjugate base or a weak base and its conjugate acid, usually in the form of a salt of the conjugate base or acid. Buffers have the property that a small change in the amount of strong acid or strong base added to them results in a much larger change in pH. The resistance of a buffer solution to pH change is due to the fact that the process of adding acid or base to the solution is slow compared to the rate at which the pH changes. In addition to this buffering action, the inclusion of the conjugate base or acid also slows the process of pH change by the mechanism of the Henderson–Hasselbalch equation. Buffers are most commonly found in aqueous solutions.

01:50

Calculate the equilibrium …

Hello. So today we're going to be looking at a situation well. We've taken 2.5 grams of copper sulfates and dissolved it into 900 milliliters of 0.3 mil aren t ammonium. So what exactly is going to happen? Well, copper two plus tends to make complex ions with ammonium, so we're going to see copper two plus ammonium and a complex ion in solution. So first, let's right the equation for the complex ion informations. We have copper two plus in solution and we have ammonium and we're forming copper with Ford Ammonia. Mum's touch to it. Two plus. So I'm gonna have need tohave before here. And so here we have it. Now let's form a nice table So we know the concentration of ammonium already ammonia. We have 0.3 mil arat e. We don't have any of the complex ion yet, so now let's try to find the concentration of the copper. So we have 2.5 grams of copper sulfate, So divide by the molar mass which is 159.609 grams Permal off copper sulfate. So every mole off copper sulfate is one mole of copper two plus I owns and then we divide by the volume we have 900 mil leaders. That's 9000.9 leaders. So we will see that our concentration of copper is 0.174 So 0.174 And so we're gonna have used up one mole of copper. Four moles of ammonium give one mall of the complex ion. So this looks pretty complex right now. So let's try to figure out a way to simplify this. And one way to simplify things is to think of in terms of limiting reagents. So the constant information the include every in constant for this reaction is 5.0 times 10 to the 13th which is really, really big Slam means that this will go almost to completion sort of. But let's take a look at our starting materials. Well, the concentration of ammonia is a lot larger than the concentration of copper, so let's consider that copper is are limiting re agent, so almost all of the copper will react with ammonium to give the complex ion. So then the concentration of our complex ion will be the concentration of copper at the start, which is 0.0 174 polarity. Now let's take a look at Not the Constitution of ammonia, well, ammonia. It takes if a zero point almost 0.174 moles of copper has reacted. That takes four times that amount, so that would be so. It would be 0.3 minus 0.174 times four, which is 0.174 times four is 0.696 Subject that from 3.3 and we get zero point 23 Similarity. So now we have the concentration of ammonium at, um, ammonia and the complex ion. So now let's find the very minuscule concentration of copper two plus that is left. So the constant, the equilibrium constant for this reaction would be the concentration of the complex Ryan over the concentration of ammonium to the fourth. Because remember has a four in front and then the Constitution of the copper two plus so as mentioned before, our great constant is 5.0 times 10 to the 13. The complex concentration will complex ion is 0.174 The concentration off ammonia is 0.23 to the fourth, then is just the construction of covered us. So how about way? Bring the silver to that side? So we'll have 5.0 times 10 to the 13th times 100.23 to the fourth, So 0.23 to the fourth is 0.28 I'm going to multiply that by five times, tend to the 13 so we will get 1.4 times 10 to the 11th. Copper two plus is equal to 0.174 No divide. They will see that we get 1.2 times 10 to the negative 13 polarity, which is very, very small, so you can see that it has. Essentially, this is essentially zero. It's like very, very small. So it's like almost essentially zero, and this has gone essentially to completion almost because this is a very

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