Math / Algebra

QuestionComplete the $K_{\text {sp }}$ expression for $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ . $K_{\mathrm{sp}}=8.46 \times 10^{-12}=$

Studdy Solution

STEP 1

What is this asking? We need to write the solubility product expression for silver carbonate ( $\text{Ag}_2\text{CO}_3$ ). Watch out! Don't forget to use the correct coefficients from the balanced dissolution equation, they're super important!

STEP 2

1. Write the balanced dissolution equation.
2. Write the $K_\text{sp}$ expression.

STEP 3

Let's start by writing out the balanced chemical equation for the dissolution of solid silver carbonate ( $\text{Ag}_2\text{CO}_3$ ) in water.
When $\text{Ag}_2\text{CO}_3$ dissolves, it breaks apart into its ions: silver ions ( $\text{Ag}^{+}$ ) and carbonate ions ( $\text{CO}_3^{2-}$ ).

STEP 4

The balanced equation looks like this: $\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2\text{Ag}^{+}(aq) + \text{CO}_3^{2-}(aq)$ See how we get two silver ions for every one carbonate ion?
That 2 is going to be super important in our next step!

STEP 5

Now, we can write the $K_\text{sp}$ expression.
Remember, the $K_\text{sp}$ is just the product of the concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced equation.
It tells us how much of the solid can dissolve!

STEP 6

From our balanced equation, we see that the coefficient of $\text{Ag}^{+}$ is 2, and the coefficient of $\text{CO}_3^{2-}$ is 1.
So, our $K_\text{sp}$ expression is: $K_\text{sp} = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}]$ We're raising the silver ion concentration to the power of 2 because we have two silver ions in the balanced equation!

STEP 7

So, the completed $K_\text{sp}$ expression is: $K_\text{sp} = 8.46 \times 10^{-12} = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}]$

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QuestionComplete the $K_{\text {sp }}$ expression for $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ . $K_{\mathrm{sp}}=8.46 \times 10^{-12}=$

Studdy Solution

STEP 1

What is this asking? We need to write the solubility product expression for silver carbonate ( $\text{Ag}_2\text{CO}_3$ ). Watch out! Don't forget to use the correct coefficients from the balanced dissolution equation, they're super important!

STEP 2

1. Write the balanced dissolution equation.
2. Write the $K_\text{sp}$ expression.

STEP 3

Let's start by writing out the balanced chemical equation for the dissolution of solid silver carbonate ( $\text{Ag}_2\text{CO}_3$ ) in water.
When $\text{Ag}_2\text{CO}_3$ dissolves, it breaks apart into its ions: silver ions ( $\text{Ag}^{+}$ ) and carbonate ions ( $\text{CO}_3^{2-}$ ).

STEP 4

The balanced equation looks like this: $\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2\text{Ag}^{+}(aq) + \text{CO}_3^{2-}(aq)$ See how we get two silver ions for every one carbonate ion?
That 2 is going to be super important in our next step!

STEP 5

Now, we can write the $K_\text{sp}$ expression.
Remember, the $K_\text{sp}$ is just the product of the concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced equation.
It tells us how much of the solid can dissolve!

STEP 6

STEP 7

So, the completed $K_\text{sp}$ expression is: $K_\text{sp} = 8.46 \times 10^{-12} = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}]$

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QuestionComplete the Ksp K_{\text {sp }}Ksp ​ expression for Ag2CO3\mathrm{Ag}_{2} \mathrm{CO}_{3}Ag2​CO3​. Ksp=8.46×10−12=K_{\mathrm{sp}}=8.46 \times 10^{-12}=Ksp​=8.46×10−12=

Studdy Solution

STEP 1

What is this asking? We need to write the solubility product expression for silver carbonate (Ag2CO3\text{Ag}_2\text{CO}_3Ag2​CO3​). Watch out! Don't forget to use the correct coefficients from the balanced dissolution equation, they're super important!

STEP 2

1. Write the balanced dissolution equation.2. Write the KspK_\text{sp}Ksp​ expression.

STEP 3

STEP 4

STEP 5

Now, we can **write the KspK_\text{sp}Ksp​ expression**.Remember, the KspK_\text{sp}Ksp​ is just the product of the concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced equation.It tells us how much of the solid can dissolve!

STEP 6

STEP 7

So, the completed KspK_\text{sp}Ksp​ expression is: Ksp=8.46×10−12=[Ag+]2[CO32−] K_\text{sp} = 8.46 \times 10^{-12} = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}] Ksp​=8.46×10−12=[Ag+]2[CO32−​]

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QuestionComplete the $K_{\text {sp }}$ expression for $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ . $K_{\mathrm{sp}}=8.46 \times 10^{-12}=$

What is this asking? We need to write the solubility product expression for silver carbonate ( $\text{Ag}_2\text{CO}_3$ ). Watch out! Don't forget to use the correct coefficients from the balanced dissolution equation, they're super important!

1. Write the balanced dissolution equation.
2. Write the $K_\text{sp}$ expression.

Now, we can write the $K_\text{sp}$ expression.
Remember, the $K_\text{sp}$ is just the product of the concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced equation.
It tells us how much of the solid can dissolve!

So, the completed $K_\text{sp}$ expression is: $K_\text{sp} = 8.46 \times 10^{-12} = [\text{Ag}^{+}]^2 [\text{CO}_3^{2-}]$