Really basic question in ionic equilibrium

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SUMMARY

This discussion focuses on solving problems related to ionic equilibrium and electrochemical cells in a grade 12 chemistry context. The key formula for calculating the H+ ion concentration from acetic acid (CH3COOH) is derived from the dissociation constant (Ka = 1.8 x 10-5). For the electrochemical cell, participants are guided to identify half-reactions for nickel and silver electrodes, with standard reduction potentials (Eo) provided as +0.80 V for Ag+ and -0.24 V for Ni2+. The discussion emphasizes the importance of equilibrium expressions and the concept of degree of dissociation.

PREREQUISITES
  • Understanding of acid dissociation constants (Ka)
  • Knowledge of equilibrium expressions in chemistry
  • Familiarity with electrochemical cells and half-reactions
  • Ability to interpret standard reduction potentials (Eo)
NEXT STEPS
  • Learn how to derive the H+ concentration using the formula Ka = [H+][A-]/[HA]
  • Study the concept of degree of dissociation (α) in weak acids
  • Explore examples of electrochemical cell calculations and net reactions
  • Review standard reduction potential tables for various half-reactions
USEFUL FOR

Students in high school chemistry courses, particularly those studying ionic equilibrium and electrochemical cells, will benefit from this discussion. It is also valuable for educators seeking to clarify these concepts for their students.

m0286
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Hello,
I am am working on my grade 12 chemistry course ALMOST DONE YEAHH!
I am stuck ona few questions, and I am hoping to get some help even if you just tell me where to begin on solving these.
1) If the [tex]Ka=1.8 * 10^{5}[/tex] for acetic acid ( ie. [tex]CH_{3}COOH[/tex]), what is the [tex]H^{+}[/tex] ion concentration in a solution of this acid, if 1.2 grams of acid are dissolved in 1.0L of solution.
Is there a formula i should be using to solve this question I can't seem to figure out what I am supposed to be doing.

I know I am going ot get told, I have to show my work so far but that's just the problem I have no clue where to begin, so if someone could just help me with that, I am not asking for you to do the question for me.

2. An electrochemical cell is constructed by placing a nickel electrode into a 1.0 M [tex]NiSO_{4}[/tex] solution, and a silver electrode into a 1.0 M [tex]AgNO_{3}[/tex] solution, and then joining them with a salt bridge to complete the circuit.
a) give the anode half reaction and the E[tex]^{o}[/tex] value.
b) give the cathode half reaction and the E[tex]^{o}[/tex] value.
c) Give the net reactiona nd its E[tex]^{o}[/tex] value.
What I am able to find out is The Eo value of Ag+ is +0.80 ans the Eo value of Ni2+ is -0.24> But other than this I am completely clueless as where to solve the question
PLEASE HELP!
 
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1)
This is a really basic question in ionic equilibirium. This is an example of dissociation of an acid in water.
Let me give you the general case.
Consider an acid HA which when dissolved in water partially ionizes into H+ and A- ions. There are H+ ions, A- ions and unionized HA in equilibirium in the solution.
Apply the http://www.science.uwaterloo.ca/~cchieh/cact/c123/massacti.html to this equilibirium.
You get,
[tex]K_a = \frac {[H+][A-]}{[HA]}[/tex]

Where [H+], [A-], [HA] are the equilibirium concentrations and [tex]K_a[/tex] is the dissociation constant of the acid HA.
Now, how do you find the H+ concentration from this?

You know the initial concentration of the acid because it is given in the question. Let the degree of dissociation of the acid be [tex]\alpha[/tex].

The degree of dissociation is a fraction of the total number of moles of an acid or base or electrolyte that dissociates into ions in an aqueous solution when equilibrium is reached.

Here, the total moles of the acid present initially is known. Call it as [itex]c[/itex]
So, from the above definition, the number of moles of ions of H+ present is [itex]c\alpha[/itex]. Similarly the number of moles of ions of A- present is also [itex]c\alpha[/itex].
So how many moles of unionized HA are present at equilibirium? From this, can you find the concentration of H+? Can you also post and show the general formula to find the concentration of H+ and the value of alpha in terms of 'c' and 'Ka'?

Note: The value of Ka for Acetic acid is 1.8 x 10^(-5) and not 1.8 x 10^(+5)
 
Last edited:


Hi there! It's great to hear that you're almost done with your chemistry course. Ionic equilibrium can be a tricky topic, but I'm sure with some guidance, you'll be able to solve these questions.

For the first question, you're given the Ka value for acetic acid and asked to find the H+ ion concentration in a solution. Remember that Ka is the equilibrium constant for the dissociation of an acid, so you'll need to set up an equilibrium expression using the formula for Ka. Then, you can plug in the given values and solve for the H+ ion concentration. If you're still stuck, try looking up examples or practice problems on Ka and acid dissociation to get a better understanding.

For the second question, you're dealing with electrochemical cells and half reactions. Remember that the anode is where oxidation occurs and the cathode is where reduction occurs. So, you'll need to identify which half reactions correspond to each electrode. The Eo value is the standard reduction potential, which you can find in a table. Once you have both half reactions and their Eo values, you can use them to calculate the net reaction and its Eo value. Again, if you're still unsure, try looking up examples or practice problems on electrochemical cells to get a better understanding.

I hope this helps you get started on these questions. Remember to always show your work and ask your teacher or classmates for help if you're still stuck. Keep up the good work and good luck with the rest of your course!
 

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