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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

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  1. Jusaquikie
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so 3/.5

  6. Jusaquikie
    • 2 years ago
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    the answer is .370 M just not getting it.

  7. aaronq
    • 2 years ago
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    yeah thats right

  8. aaronq
    • 2 years ago
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    so 3/.5 this is wrong, but the answer is 0.37 M

  9. Jusaquikie
    • 2 years ago
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    i understand now thank you

  10. aaronq
    • 2 years ago
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    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
    • 2 years ago
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    okay good stuff

  12. Jusaquikie
    • 2 years ago
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    .08 + .105 / .5

  13. Jusaquikie
    • 2 years ago
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    thank you for your time it is appreciated.

  14. aaronq
    • 2 years ago
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    no problem

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