Question

Molarity question?

Consider the precipitation reaction of CuCl2 with NaOH:

CuCl2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaCl(aq)

A. What is the theoretical yield(in moles) of copper (II) hydroxide when 100mL of 0.125M CuCl2 is combined with 100 mL of 0.230M NaOH?

I got 0.0115 mol Cu(OH)2

B. What is the concentration (M) of each anion in solution after the reaction is complete?

I got 0.125 M for Cl. I know that the M for OH is 0, but I don't know why.

C. When this reaction takes place in a coffee-cup calorimeter, the temperature increases from 23 degrees Celsius to 33.3 degrees Celsius. What is `Delta`H for the precipitation of copper(II) hydroxide (kJ/mol)? Assume that the solution has the same specific heat and density as water.

Answer 1

See below.

Explanation:

Part A: Correct answer.

Part B:

Step 1: Find the mols of `NaCl` produced from the limiting reagent `NaOH`

`"0.0230 moles NaOH"*(("2 moles NaCl")/("2 moles NaOH"))="0.0230 moles NaCl produced"`

Step 2: Calculate molarity of `Cl^(-)` anions.

There will be 1 mole of `Cl^(-)` for every 1 mole of `Na^(+)`, so we have `"0.0115 moles" Cl^(-)`

Plug into molarity equation:

`M=("0.0115 moles"/"0.200 liters")=0.0575M`

Step 3: There will be no concentration of `OH^(-)` ions because they are in a solid compound, not dissolved in solution. They will no longer be anions in a solid, neutral compound.

Thus, molarity of `OH^(-)` is 0.

Part C:

Step 1: Calculate `q` for the heat the water absorbed , not the heat released by the reaction.

`q=s*m*DeltaT`

We know `s` for water is `4.184 J/(g*K)`, `m` will be calculated using a density of `"1 gram"/"mL"`, and `DeltaT` will be found by `"final T"-"initial T"`.

Step 2: Let's plug in the numbers.

`q=4.184*(200 * 1)*(33.3-23)`

Step 3: Simplify.

`q=4.184*200*10.3="8,619 J"="8.619 kJ"` (not using the correct number of sig figs)

Step 4: Convert to heat released.

This number is the amount of heat the water absorbed during the reaction for `"0.0115 moles of" CuCl_2`. To turn this into the heat released by the reaction, we simply make it exothermic `q="-8.619 kJ"/"0.0115 moles"`.

Step 5: Convert to `"kJ"/"mol"`

Now that we have the heat released, we need to convert this value into `"kJ"/"mol"`.

`-8.619*(1/0.0115)=-749 "kJ"/"mol"`

Step 6: All done!

I hope this helps!