Question

Indicate the concentration of each ion or molecule present:
a mixture of 48.0 mL of 0.11 M KClO3 and 28.0 mL of 0.29 M Na2SO4. Assume the volumes are additive.
a) [K+]
b) [ClO3-]
c) [SO42-]
d) [Na+]

We haven't taken notes on this particular type of problem yet and I can't find tutorials on the internet. Do I have to find the number of moles of each substance, then add them or something similar? I'd appreciate if someone could show me how to do at least one so I can figure out the rest on my own.

Answer 1

The problem asks for concentration and probably that means M or mols/L (or millimols/mL). So you want to find the mols of each ion, divide by total volume and that will be M. I prefer to find millimols and divide by mL.

For 48 mL 0.11 M KClO3 you have 48 x 0.11 = 5.28 millimols. The total volume is 48.0 mL + 28.0 mL =76.0 mL. So (KClO3) is 5.28mmols/76.0 mL = 0.06947 M which I would round to 0.0695 to three significant figures. Now that you know the (KClO3) you know (K^+) = 0.0695 M and (ClO3^-) = 0.0695 M. You do Na2SO4 the same way. Remember, though, after you have (Na2SO4), the (Na^+) will be 2x (Na2SO4) because there are 2 Na^+ per molecule Na2SO4.

There is another way to do this that I think is easier. You're mixing two substances and each dilutes each other; i.e., the 0.11M KClO3 is diluted from 48.0 mL to 76.0 mL so the new concentration is 0.11 M x (48.0/76.0) = 0.0695M with (K^+) = (ClO3^-) = 0.0695 M.
For Na2SO4, it is
0.29 M x (28.0/76.0) = 0.1068 which I would round to 0.107 M. That gives you (Na^+) = 2*0.107 = ? and (SO4^2-) = 0.107 M.