Question

400 cc of 0.2M cacl2 reacts with 600cc 0.15M na2so4 solution. What is the Molarity of so4 2- in final solution?

Answer 1

We interrogate the reaction...`CaCl_2(aq) + Na_2SO_4(aq) rarr CaSO_4(s)darr + 2NaCl(aq)`

Explanation:

Note that calcium sulfate is pretty insoluble, and here it is assumed that it all precipitates. Had the question wanted you to consider the solubility of this salt it WOULD have quoted a `K_"sp"` value.

`"Moles of calcium ion"=400*mLxx10^-3*L*mL^-1xx0.20*mol*L^-1=0.080*mol`

`"Moles of sulfate ion"=600*mLxx10^-3*L*mL^-1xx0.15*mol*L^-1=0.090*mol`

And thus there is EXCESS sulfate ion in the NEW solution...i.e. `0.01*mol` in the additive volume of `1000*mL-=1000*cm^3-=1*L`..

`[SO_4^(2-)]=(0.01*mol)/(1.0*L)=0.010*mol*L^-1`