Explore
For Enterprise
Join for Free
Log In

Explorar
Buscar
For Enterprise
Inicia Sesión
Únete de forma gratuita

4.12.b Aqueous Equilibria Worked Example

Loading...

Advanced Chemistry

Universidad de Kentucky

4.7 (444 calificaciones) | 52K estudiantes inscritos

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

Revisiones

4.7 (444 calificaciones)

5 stars
333 ratings
4 stars
89 ratings
3 stars
16 ratings
2 stars
4 ratings
1 star
2 ratings

KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

De la lección

Aqueous Equilibria

This unit continues and expands on the theme of equlibria. You will examine buffers, acid/base titrations and the equilibria of insoluble salts.

4.01 Buffers and the Common Ion Effect10:01

4.02 pH of Buffer Solutions14:04

4.02a Aqueous Equilibria Worked Example 13:58

4.03 Buffer Action25:21

4.03a Aqueous Equilibria Worked Example 25:22

4.03b Aqueous Equilibria Worked Example 35:11

4.04 Buffer: Preparation and Capacity8:45

4.05 Strong Acid - Strong Base Titration17:20

4.06 Titrations Involving Either a Weak Acid or a Weak Base36:45

4.06 Part 1 Aqueous Equilibria Worked Example3:46

4.06 Part 1.a - Aqueous Equilibria Worked Example4:06

4.06 Part 1.b - Aqueous Equilibria Worked Example4:55

4.06 Part 1.c - Aqueous Equilibria Worked Example4:21

4.06 Part 1.f Aqueous Equilibria Worked Example8:01

4.06 Part 1.h Aqueous Equilibria Worked Example3:57

4.06 Part 2.a Aqueous Equilibria Worked Example8:25

4.06 Part 2.b Aqueous Equilibria Worked Example5:53

4.07 Polyprotic Acid Titrations5:53

4.08 Indicators10:19

4.09 Solubility Equilibria4:50

4.10 Molar Solubility and the Solubility Product10:37

4.10.a Aqueous Equilibria Worked Example2:20

4.11 Molar Solubility and the Common Ion Effect7:20

4.11 2.b - Aqueous Equilibria Worked Example4:42

4.11 2.c Aqueous Equilibria Worked Example3:45

4.12 The Effect of pH on Solubility7:07

4.12.b Aqueous Equilibria Worked Example3:42

4.12c Aqueous Equilibria Worked Example3:30

4.13 Precipitation Reaction and Selective Precipitation15:03

4.13a Aqueous Equilibria Worked Example9:05

Impartido por:

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

Transcripción

In the previous learning objective, we worked through part a where we determine the molar solubility of copper hydroxide in just water. In this problem, we're going to determine the molar solubility in a solution that's buffered at pH of 9. Now this is a type of common ion effect problem. But it's seems a little strange by then talking about us having a solution that's buffered at a pH of 9. But let's beginning by writing the reaction of the copper hydroxide and water. Okay, what does it do? It dissolves so this is the dissolution into its ions. So this is how we did part A previously. Where it changes is in the I line, because we're not dissolving it in water, we're dissolving it in a buffer with a pH of 9. So let's see what the hydroxide concentration would be if the pH were 9. Well we know that if the PH is 9, then the POH is 14-9. And that would be 5. If we wanted to know the Concentration, that would be 10 to the -5 or, 1.0 x 10 to the -5. So, when we dump some of the solid into the solution it's not into water, but there is no copper present, but there is already some hydroxide present. So we have some of these ion present. Now we will have -s, which would be once we find s that's the molar solubility. This is how much we dissolve, we have s on the side and we'd have +2s on this side. So there's still some of the solid sitting at the bottom, there'd be s here for the copper 2+, and I'm going to put 1.0 x 10 to the -5 and that's all, and not add +s. Why is that? Well, because it tells me it's buffered to that pH. So it is resisting change to pH, it is maintaining this hydroxide concentration. Now we're ready to calculate the s value so we take our 2.2 x 10 to the -20, and that would be the concentration of copper times the concentration of hydroxide squared. We plug in values that we know. Which we have s for this and we have (1.0 x 10 to -5) squared for this, and that will give me s = 2.2 x 10 to -20 divided by 1.0 x 10 to -10. Because -5 squared is to -10. This will give me a solubility of 2.2 x 10 to the -10. That will be the concentration of the copper hydroxide that would dissolve. Now, what did we learn in this unit? We learned that if you have a solution that contains a common ion, it will always decrease the solubility of that salt in the solution. What I've written up here, what we obtained for the value of copper hydroxide solubility in pure water. 1.9 x 10 to the -7. This is one 2.2 x 10 to the -10. That is lower. It has decreased the solubility by having a common ion present in solution.

Explora nuestro catálogo

Inscríbete de manera gratuita y obtén recomendaciones personalizadas, actualizaciones y ofertas.

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.

© 2019 Coursera Inc. Todos los derechos reservados.

Consíguelo en Google Play

Coursera
Acerca de
Liderazgo
Empleo
Catálogo
Certificados
Certificados MasterTrack™
Grados
For Enterprise
For Government
Para el campus

Comunidad
Learners
Partners
Developers
Beta Testers
Translators

Conectar
Blog
Facebook
LinkedIn
Twitter
Instagram
Youtube
Blog de Tecnología

Más
Términos
Privacidad
Ayuda
Accesibilidad
Prensa
Contacto
Directorio
Afiliados