Calculate the concentration of clo2− at equilibrium if the initial concentration of hclo2 is 2. 25×10^−2 m.

Chemistry

1 answer:

jek_recluse [69]1 year ago

6 0

The concentration of ClO₂⁻ at equilibrium if the initial concentration of HClO₂ is 0.0654.

<h3>What is concentration?</h3>

The concentration of any substance is the quantity of that substance in per square of the space or container.

The reaction is

HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻

The pH is 0.454 M

Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]

2. 25 × 10⁻² m = [x][x] / 0.454-x]

2 + 0.011 - 0.004994 = 0

solve the quadratic equation

x = 0.0654 = [H3O+] = [ClO2-]

pH = -log (H3O+)

pH = -log(0.0654)

pH = 1.2

equilibrium concentrations of

[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M

[ClO2- ] = x = 0.0654

Thus, the equilibrium concentrations is 0.0654.

To learn more about concentration, refer to the link:

brainly.com/question/16645766

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<u>First of all, what is electrolysis?</u>

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→ Positively charged ions move to the negative electrode. They receive electrons and are <em>reduced</em>. The positive ions move towards the negative electrode because they want to cancel each other out.

→ Negatively charged ions move to the positive electrode. They lose electrons and are <em>oxidised</em>. The substance that is broken down is called the electrolyte <em>(an electrolyte is just a liquid or solution that can conduct electricity)</em> . The negative ions move towards the positive electrode because they want to cancel each other out.

<h3>Cathode = Negative electrode</h3><h3>Anode = Positive electrode</h3>

Metal ions form at the cathode and non-metal ions form at the anode

How I remember if an element is <em>oxidised</em> or <em>reduced</em> is by remembering OIL RIG

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<h2><em><u>The answer to your question</u></em></h2>

1) The first step would be to clean the metal ring and sand it down because when the metal atoms from the electrolyte are deposited onto the ring, they will form a weak bond and they may simply 'fall' off. Also this could affect conductivity and the whole experiment. The more things you do accurately now, the more accurate your result will be.

2) You want to put the solution you are given in to the tank your going to be using.

3) This is basically the main part, you want to set up the circuit, I have attached a diagram at the bottom to show you the circuit. The copper rod will be the anode and the metal ring will be a cathode (ignore the elements).

4) Now turn on the circuit and you will start to see the solution spilt with the the solution now being split some going to the anode and some going the cathode.

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Hope this helps :)