Answer to Question #143357 in General Chemistry for Nicole

Question

A 24.0 mL sample of 0.163 M Al(NO3)3 sample is mixed with a 19.0 mL sample of 0.167 M K2CO3. What is the concentration of Al3+ after this reaction finishes?

Answer 1

Q143357

A 24.0 mL sample of 0.163 M Al(NO3)3 sample is mixed with a 19.0 mL sample of 0.167 M K2CO3. What is the concentration of Al3+ after this reaction finishes?

Solution:

Step 1: Write the balanced equation of Al(NO₃)₃ and K₂CO₃ .

Al(NO₃)₃ and K₂CO₃ will undergo a double displacement reaction.

Al(NO₃)₃ dissociates to give Al ⁺³ and NO₃^- ions.

K₂CO₃ dissociates to give K⁺ and CO₃^2- ions.

The interchange of cations and anions will take place in this reaction.

Al⁺³ will combine with CO₃^2- and K⁺ will combine with NO₃^-

Al⁺³ and CO₃ ^2- forms Al₂ (CO₃)₃

K⁺ and NO₃ ^- will form KNO₃ ;

So the reaction will be

Al(NO₃)₃ + K₂CO₃ . ----------> Al₂ (CO₃)₃ + KNO₃ ;

Now we balance this reaction,

balance Al on both the side we have

2 Al(NO₃)₃ + K₂CO₃ . ----------> Al₂ (CO₃)₃ + KNO₃ ;

balance NO₃ on both the side.

On left side there is 6 NO₃^- and on right side there is 1 NO₃^- .

We can balance NO₃ by writing 6 in front of KNO₃

2 Al(NO₃)₃ + K₂CO₃ . ----------> Al₂ (CO₃)₃ + 6 KNO₃

next we will balance K on both the side.

There is 2 K on left side and 6 K on right side.

After writing 3 in front of K₂CO₃ we can balance K.

2 Al(NO₃)₃ + 3 K₂CO₃ . ----------> Al₂ (CO₃)₃ + 6 KNO₃

Now the reaction is balanced for all the elements.

Step 2 : Find the moles of Al(NO₃)₃ and K₂CO₃ . Using the given volume and concentration of

Al(NO₃)₃ and K₂CO₃

Molarity = moles of solute/ Volume in ‘L’

we are given 24.0 mL sample of 0.163 M Al(NO₃)₃ and 19.0 mL sample of 0.167 M K₂CO₃

Convert 24.0mL and 19.0mL to ‘L’

volume of Al(NO₃)₃ in ‘L’ = 24.0mL * 1L/1000mL = 0.024 L

volume of K₂CO₃ in ‘L’ = 19.0mL * 1L / 1000mL = 0.019 L

For Al(NO₃)₃ , we have

0.163 M = moles of Al(NO₃)₃ /0.024 L

multiply both the side by 0.024L we have

0.163 mol/L * 0.024L = moles of Al(NO₃)₃ / 0.024 L * 0.024L ; (∵ M = mol/L)

0.003912 moles = moles of Al(NO₃)₃

So, moles of Al(NO₃)₃ = 0.003912 moles.

For K₂CO₃ , we have

0.167 M = moles of K₂CO₃ ,/ 0.019 L

multiply both the side by 0.019L we have

0.167 mol/L * 0.019L = moles of K₂CO₃,/0.019 L * 0.019L ; (∵ M = mol/L)

0.003173 moles = moles of K₂CO₃

So, moles of K₂CO₃ = 0.003173 moles.

Step 3 : To find Limiting reactant.

Consider the reaction

2 Al(NO₃)₃ + 3 K₂CO₃ . ----------> Al₂ (CO₃)₃ + 6 KNO₃ ;

We will find the moles of Al(NO₃)₃ that will react with 0.003173 moles of K₂CO₃ .

Use the mole ratio of Al(NO₃)₃ and K₂CO₃ from the reaction. The mole ratio is 2 : 3.

Moles of Al(NO₃)₃ that will react with given moles of K₂CO₃ .

= 0.003173 moles of K₂CO₃ * 2 mol of Al(NO₃)₃ / 3 mol of K₂CO₃

= 0.002115 mole of Al(NO₃)₃ ;

So only 0.002115 mole of Al(NO₃)₃ will react with K₂CO₃ .

K₂CO₃ is the limiting reactant and Al(NO₃)₃ is the excess reactant.

Step 4 : To find the moles of excess reactant left

moles of Al(NO₃)₃ left = moles of Al(NO₃)₃ present in the solution - moles of Al(NO₃)₃ reacted

= 0.003912 moles - 0.002115 mole.

= 0.00180 moles .

Step 5 : To find the concentration of Al⁺³ after the reaction is complete.

The final volume of the solution in ‘L’ = 0.024L + 0.019 L = 0.043 L;

moles of Al(NO₃)₃ left after reaction = 0.00180 moles ;

Each Al(NO₃)₃ can give 1 Al ⁺³

So, moles of Al ⁺³ left in the solution will also be = 0.00180 mole

Molarity of Al⁺³ = 0.00180 mol Al⁺³ / 0.043 L

Molarity of Al⁺³ = 0.04186 mol/ L

In question we are given all the quantities in 3 significant figure. So our final answer must also be in

3 significant figure.

Hence the concentration of Al⁺³ after completion of reaction is 0.0419 M ;