# Concentration of Ions at Equilibrium

• Revengeance
In summary, the problem involves finding the concentrations of all ions in a solution at equilibrium after mixing two solutions. The balanced equation is Cu(NO3)2 + 2KOH -> Cu(OH)2 + 2K(NO3) and the Ksp for Cu(OH)2 is given. To solve, first determine the limiting reagent or excess ion, then use Ksp to calculate the concentration of the other ion. The concentration of Cu2+ is 0.187 M and the concentration of OH- can be found using Ksp = [OH]^2[Cu].
Revengeance

## Homework Statement

Find the concentrations of all ions in the solution at equilibrium after 0.213 L of 0.433 M aqueous Cu(NO3)2 solution is mixed with 0.213 L of 0.120 M KOH solution. Ksp for Cu(OH)2 is 1.1E-15.

## Homework Equations

n=m/M

[Cu][OH]^2 = Ksp .. ?

## The Attempt at a Solution

ok so for the solution what i did was write out the balanced equation

Cu(NO3)2 + 2KOH -> Cu(OH)2 + 2K(NO3)

So we are told that the Cu(OH)2 = Ksp, so that means the spectator ions NO3 and K are going to be, their moles over the new volume. So i solved for them already and had received 0.061 mol/l K 0.433 mol NO3.

For Cu2+, what i did was i took the moles of Cu2+ initially, then i found moles KOH in terms of CuNO3, and i subtracted them, then divded the moles by the total volume to get 0.190 which is the same answer they got.

Now for OH, would i do the same process as above? Where i take the original moles of OH and then subtract it from moles of CuNO3 in terms of KOH, then divide that moles by the total volume?

#### Attachments

• ksp.PNG
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Basically it is a limiting reagent problem first - check which ion (Cu2+ or OH-) is left in excess after the reaction. This will give you an easy to calculate concentration of this ion. Then use Ksp to calculate the concentration of the other ion.

Actually to be precise you should take into account both ions from the dissolution, but with so low Ksp it probably doesn't matter.

Borek said:
Basically it is a limiting reagent problem first - check which ion (Cu2+ or OH-) is left in excess after the reaction. This will give you an easy to calculate concentration of this ion. Then use Ksp to calculate the concentration of the other ion.

Actually to be precise you should take into account both ions from the dissolution, but with so low Ksp it probably doesn't matter.

Hmm... so how would i check if Cu2+ or OH- is the limiting reagent or left in excess? I already have the concentration of Cu2+ at equilibrium. So in order to calculate the OH, would i have to use Ksp = [OH]^2[Cu] ?

Revengeance said:
so how would i check if Cu2+ or OH- is the limiting reagent or left in excess?

Simple stoichiometry, google for limiting reagent

Your concentration of Cu2+ is almost OK, I got 0.187 M, no idea where the difference comes from, unless you rounded it down (but all the concentration data is given with 3SD, so this part of the answer should have 3 SD as well).

And yes, assuming you know the concentration of Cu2+ it is time to calculate OH- using Ksp.

## 1. What is the meaning of concentration of ions at equilibrium?

The concentration of ions at equilibrium refers to the relative amounts of positively and negatively charged particles in a solution when the rate of their formation is equal to the rate of their recombination, resulting in a stable concentration of ions.

## 2. How is the concentration of ions at equilibrium calculated?

The concentration of ions at equilibrium is calculated using the equilibrium constant, which is the ratio of the concentrations of products to reactants at equilibrium. It is also influenced by factors such as temperature, pressure, and the nature of the substances involved.

## 3. What is the significance of concentration of ions at equilibrium in chemical reactions?

The concentration of ions at equilibrium is crucial in determining the direction and extent of a chemical reaction. It helps in understanding the equilibrium state of a system and predicting how changes in conditions will affect the equilibrium position.

## 4. How does the concentration of ions at equilibrium affect the pH of a solution?

The concentration of ions at equilibrium has a direct impact on the pH of a solution. In acidic solutions, the concentration of hydrogen ions (H+) is higher than that of hydroxide ions (OH-), while in basic solutions, the concentration of hydroxide ions is higher. The equilibrium between these two ions determines the pH of a solution.

## 5. What factors can affect the concentration of ions at equilibrium?

The concentration of ions at equilibrium can be influenced by various factors such as temperature, pressure, and the presence of other substances in the solution. Changes in these conditions can shift the equilibrium position and alter the concentration of ions at equilibrium.

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