[govibe]

These "ion concentrations after equation" problems are a challenge for me.

"Calculate the sodium ion concentration when 60.0 mL of 8.5 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate."
OK, I first converted both the sodium carbonate and sodium bicarbonate quantities into moles... Since there are two sodium ions per sodium carbonate ion, I multipled that value by two... I then added that value with the moles of sodium ion in sodium bicarbonate, which comes out to 1.05 moles. I then divide that by .09L to get 1M, which isn't the right answer. Can someone point out what I should have done?

[Borek]

which comes out to 1.05 moles

Check your math. From your description you are approaching problem correctly, but 1.05 is about twice too much.

[govibe]

this is what I did

(.06L)(8.5M)2=1.02mol
(.03L)(1.0M)=.03mp;

1.02 +.03=1.05mol

1.05mol/.09L = 11.6666666M

What am I doing wrong?

[Borek]

Sorry, my mistake. Your result is OK (although it is not 1M you gave earlier).

Are you sure it is 8.5M? Looks like it is quite above carbonate solubility (about 3.75 M for saturated solution at 25 deg C)

[govibe]

Yes, it is 8.5M...

Apparently, the answer to this problem is 12M...

[enahs]

In your original question as you typed them your concentrations only had 2 sig figs, so your answer must be limited to 2 as well.

11 (2/3) gets rounded up to 12.