4. A 50.0-mL sample of a 1.00 M solution of CuSO4 is mixed...

4. A 50.0-mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2°C before mixing and 26.3°C after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data calculate ?H for the process CuSO4(1 M) + 2KOH(2 M) ———> Cu(OH)2(s) + K2SO4(0.5 M) Assume the specific heat and density of the solution after mixing are the same as those of pure water.

Transcript

00:01 This is sort of an interesting problem because we're told that we have 50 milliliters of 1 .00 molar of the copper 2 sulfate and 50 .0 of 2 .00 molar koh.

00:33 So this will equal 0 .050 times 1 .00 equals 0 .05 moles and 0 .050 times 2 equals 0 .10 moles.

00:59 So now if we figure out from our balanced chemical equation here 0 .10 moles koh will consume 2 to 1 koh and copper 2 sulfate.

01:19 We'll see we're at 0 .05 so we're stoichiometrically equivalent on each side.

01:34 So i have when it's telling me to do this for kilojoules per mole for the process that's sort of a weird thing because we don't have the same number of moles.

01:45 We're told to report kilojoules per mole for the process but do we want mole of koh or mole of copper 2 sulfate? i'll give you both.

02:09 It's not a big deal.

02:10 We don't have the same number of moles of each.

02:13 Okay so we do know the rest of this information here though.

02:16 Okay our initial temperature for both solutions is 20 .2 degrees celsius.

02:24 Our final temperature for both solutions is 26 .2 degrees celsius.

02:31 This will give me a delta t of 26 .2 degrees c minus 20 .2 degrees c and that'll be 6 .0 degrees c is my delta t.

02:47 We're told that our specific heat will be 4 .184 j over g degrees c.

02:55 Our density is equal to one gram per milliliter and we know that we had 50 milliliters 50 .0 milliliters of each solution equals 100 .000 milliliters total volume and 100 .00 milliliters times oops times one gram per milliliter equals 100 .0 grams and we're told that we have a calorimeter constant of lost my stuff here there it is...

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