anonymous
  • anonymous
If it requires 48.3 milliliters of 0.55 molar nitric acid to neutralize 15.0 milliliters of barium hydroxide, solve for the molarity of the barium hydroxide solution. Show all of the work used to solve this problem. Unbalanced equation: Ba(OH)2 + HNO3 yields Ba(NO3)2 + H2O
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@taramgrant0543664 @Photon336
taramgrant0543664
  • taramgrant0543664
First step: You can find the moles of nitric acid using n=cv where n is the number of moles c is the concentration (aka molarity) so 0.55M v is volume and would need to be in litres not millilitres
anonymous
  • anonymous
ok im here

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taramgrant0543664
  • taramgrant0543664
Second step: Balance chemical equation using the stoichiometric coefficients relate the HNO3 to the Ba(NO3)2 Using the ratio you can convert the moles of HNO3 to moles of Ba(NO3)2
taramgrant0543664
  • taramgrant0543664
Third Step: After using the ratio you will have the moles of Ba(NO3)2 So you can use the formula c=n/v c is the concentration that you are trying to find n is the moles you found in the previous step v is the volume in litres so the 15mL but converted
anonymous
  • anonymous
ok so the balanced equation is Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
taramgrant0543664
  • taramgrant0543664
Yes it would be
taramgrant0543664
  • taramgrant0543664
So the ratio is 2 to 1 For every 2 moles of HNO3 there is 1 mole of Ba(NO3)2 produced
anonymous
  • anonymous
ok then what do i do
taramgrant0543664
  • taramgrant0543664
SO since the ratio is 2:1 the moles you found in step 1 you divide that by 2 to get the moles of Ba(NO3)2 and then you can carry on to step 3
anonymous
  • anonymous
C=N/V 0.55M=n/v
anonymous
  • anonymous
i thought ml was volume
taramgrant0543664
  • taramgrant0543664
mL is volume but since molarity is mol/L we have to convert the millilitres into litres 1L=1000mL
anonymous
  • anonymous
but it's not 1000 ml
taramgrant0543664
  • taramgrant0543664
That's the conversion 1mL=0.001L 48.3mL=? L
anonymous
  • anonymous
0.0483L
taramgrant0543664
  • taramgrant0543664
Yes using this and the 0.55M you can find n using the n=cv formula
anonymous
  • anonymous
0.0483*0.55=0.26565
taramgrant0543664
  • taramgrant0543664
Close I think you're missing a 0 in there
anonymous
  • anonymous
0.026565
Photon336
  • Photon336
remember to add both volumes (V1 + V2) to get your final volume
anonymous
  • anonymous
huh so 0.026565 isn't the answer?
taramgrant0543664
  • taramgrant0543664
Yes so now we look at the ratio it was 2:1 so to get the moles of Ba(NO3)2 you divide the moles you just got by 2 to get the moles of Ba(NO3)2
anonymous
  • anonymous
0.026565/2=0.0132825
taramgrant0543664
  • taramgrant0543664
Yes so that would be the moles of Ba(NO3)2 So now you put that into the formula c=n/v and as @Photon336 had said you have to use the total volume so you have to add both together and convert that from millilitres to litres
anonymous
  • anonymous
so 48.3+15.0?
taramgrant0543664
  • taramgrant0543664
Yeo together that will be the final volume of the solution
anonymous
  • anonymous
48.3+15.0=63.3 ---> 0.0633
Photon336
  • Photon336
yep
anonymous
  • anonymous
so thats the final answer?
Photon336
  • Photon336
@lovingod786 they asked for concentration i think you need the number of moles too so it has to be moles/liter
taramgrant0543664
  • taramgrant0543664
Nope you have to put it into the formula first c=n/v
anonymous
  • anonymous
ok so c=0.26565/0.0633?
taramgrant0543664
  • taramgrant0543664
0.0132825/0.0633=c
anonymous
  • anonymous
0.21
taramgrant0543664
  • taramgrant0543664
yep that would be correct
Photon336
  • Photon336
m\[M _{1}V_{1} = m _{2}V_{2} \] \[m_{2} = \frac{ M_{1}V{1} }{V_{2} }\] \[m_{2}] = moles of Ba(OH)2 \[V_{2}] = final volume.
anonymous
  • anonymous
so thats the final answer then??
anonymous
  • anonymous
??
taramgrant0543664
  • taramgrant0543664
Yep
anonymous
  • anonymous
OK THX
anonymous
  • anonymous
You is a really good tutor or teacher or helper..you know whatever you clarify yourself as but you did a really good job!
anonymous
  • anonymous
Well do not did
taramgrant0543664
  • taramgrant0543664
Haha thanks! I try my best! I'm pretty bad at explaining things but this site has made me a lot better at explaining what I mean and @Photon336 is really good at explaining things too, probably one of the better people at explaining things in this section
anonymous
  • anonymous
Yeah @Photon336 is good too when I first got here they use to help me
Photon336
  • Photon336
thanks guys

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