Question

what is the final concentration of Ammonium nitrate after 250 mL of water are added to a 500 mL of a 0.175 M solution of ammonium nitrate?

Answer 1

so first write a balanced equation for this, then find how many moles there are of each species.

Answer 2

the balanced equation for ammonium nitrate is .. HNO3(aq) + NH3(g) -> NH4NO3(aq)

Answer 3

okay now find the moles of each species present initially. \(Molarity=\dfrac{n_{solute}}{L_{solution}}\)

Answer 4

wait what?

Answer 5

250 mL = 0.25 L
and 500 mL = 0.5 L

Answer 6

find moles

Answer 7

how do you find moles?

Answer 8

wait did you change the question? theres no NaClO3 anymore.

this is just a dilution: \(M_1V_1=M_2V_2\)

Answer 9

i messed the question up so i reposted it with the correct info
and what do u mean a dilution?

Answer 10

sorry for the mistake ;/

Answer 11

it's alright. A dilution means that you're making the solution LESS concentrated.

so all you need to do is use the formula above (remember that there is a new volume)

\((0.5L)(0.175 M)=(0.25 L + 0.5 L)M_2\)

solve for \(M_2\)

Answer 12

which is the new molarity

Answer 13

so then it'd be 0.75?

Answer 14

if thats what you computed then yes

Answer 15

well i just added 0.25 + 0.5

Answer 16

yeah thats the new volume, not the new molarity

Answer 17

so what do i have to do to find the molarity then

Answer 18

solve for \(M_2\)

\((0.5L)(0.175M)=(0.25L+0.5L)M_2\)

Answer 19

hmm how would i do that

Answer 20

I'm sorry I've never done a problem like this before

Answer 21

??

Answer 22

it's algebra

\((0.5L)(0.175M)=(0.25L+0.5L)m_2 \rightarrow M_2=\dfrac{(0.5L)(0.175M)}{(0.25L+0.5L)}\)

Answer 23

damn they should both say \(M_2\)