## Jusaquikie 2 years ago a 200.0 mL solution of .200 M Mg(no3)2 is mixed with a 300.0 mL solution of .350 M NaNO3 what is the concentration of the nitrate ion in the resulting solution

1. Jusaquikie

i did the folowing $M=\frac{ n }{ V }$$.2L*.2 Molarity= .04 grams$$.350L*.120 Molarity=.105=grams$ not sure where to go from here.

2. aaronq

those would be moles, not grams. M=n/V, you're saying MxL=n, so moles. you in the first Mg(no3)2 , you have 2 times NO3 right? so you have twice as many moles. in the secondm NaNO3, only once so now add them up all together and divide by the TOTAL volume, to find the concentration in terms of molarity.

3. Jusaquikie

yes i messed that step up earlier i forgot that i had to m*MM to get grams. but i'm still not sure how to do this problem i get that the total volume would be .5L but what am dividing by this? i know what there are 2 in the first and 1 in the second.

4. aaronq

they're asking you for the concentration (I'm assuming in molarity) of the NO3^- ion. so M= (total number of moles in solution)/0.5L

5. Jusaquikie

so 3/.5

6. Jusaquikie

the answer is .370 M just not getting it.

7. aaronq

yeah thats right

8. aaronq

so 3/.5 this is wrong, but the answer is 0.37 M

9. Jusaquikie

i understand now thank you

10. aaronq

so from the first you have 0.2M (0.2L) = 0.04 moles x 2 = 0.08 moles from the second you have 0.35M(0.3 L) = 0.105 moles M=(0.08+0.105)/0.5L

11. aaronq

okay good stuff

12. Jusaquikie

.08 + .105 / .5

13. Jusaquikie

thank you for your time it is appreciated.

14. aaronq

no problem

Find more explanations on OpenStudy