Including activity coefficients find cu2 in saturated cuio32...

Including activity coefficients, find [Cu2+] in saturated Cu(IO3)2 in 0.175 M NaIO3. The solubility of Cu(IO3)2 is small, so the concentration of IO3− is 0.175 M from NaIO3. The Ksp for Cu(IO3)2 is 1.4 x 10−7, the hydrated radii (α) are 600 pm for Cu2+ and 450 pm for IO3−, and the ionic strength of the solution (μ) is 0.175 M (YOU do not have to calculate this).

Transcript

00:01 Hello everyone that's all this question in this question we have a reaction that is copper hydroxide dissociate into copper plus 2 plus 2 oh h negative in equestate and the ksp for this reaction is 4 .8 into 10 multiplied by 10 this to power minus 20 and we have another reaction c ui o 3 hold twice dissociates into copper plus 2 plus i .o3 negative ions and ksp for this reaction is 6 .94 multiplied by 10 is to power minus 8.

00:46 This is given and we have to find out ksp for the third reaction.

00:53 So, let us understand ksp for the first one is equals to if we write in the formula, it will.

01:04 Will be products, concentration of copper plus 2 and concentration of oh negative ions raised to the power 2 divided by copper hydroxide but because it is in solid state it is taken as 1.

01:21 So, this value is equal to 4 .8 multiplied by 10 times to bar minus 20.

01:27 And ksp for the second one, similarly only the product ions will be considered as the the reactant is solid state, so its concentration is taken as 1.

01:41 And this is equal to 6 .94 multiplied by 10 to power minus 8.

01:47 Now we have another reaction for whose case we have to find out, that is cyo3 holds twice in solid state, plus 2.

01:58 Oh negative ion in aqua state gives copper hydroxide solid state, plus to i .o3 negative aqua state.

02:10 So equilibrium constant k is equals to for this reaction i .o3 minus whole square divided by oh negative ion whole square.

02:32 So now multiplying and dividing both the numerator and denominator denominator by copper plus two.

02:47 So we will get k is equals to let us write it i .03 minus whole square copper plus 2 and oh h negative whole square copper plus 2 now we can write it as k s b 2 and this one is k sb 1 so k can be written as 6 .94 multiplied by 10x to 4 minus 8 divided by 4 .8 multiplied by 10 to par minus 20.

03:21 So, k is equals to 1 .45 multiplied by 10 raised to par 12.

03:26 So, this is answer to the first part of the question.

03:31 The another question here is the question in which it is given that a reaction, copper ion and i .o3 negative ions are formed from, formed from cu i...

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