Jhannybean
  • Jhannybean
A student needs to standardize her NaOH she made by diluting \(\sf 10~mL\) of \(\sf 6~M~ NaOH\) with \(1000~mL\) of \(\sf H_2O\). Calculate the Molarity of NaOH solution of \(\sf 0.2001~g~KHP\); \(\sf MM_{KHP} ~=~ 204.23~\frac{g}{mol}\) ; that requires \(\sf 17.02~mL\) of the diluted \(\sf NaOH\)
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
@Shalante @Photon336 @aaronq
Jhannybean
  • Jhannybean
the \(\sf 1000~ml~H_2O\) is just extra information
anonymous
  • anonymous
you need to know the molar ratio's of NaOH and KHP

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anonymous
  • anonymous
most likely its 1:1
anonymous
  • anonymous
now C=n/V \[C _{NaOH}=\frac{ n _{NaOH} }{ V _{NaOH} }\] But since mole ration is 1:1 \[n _{NaOH}=n _{KHP}\] Therefore \[C _{NaOH}=\frac{ n _{KHP} }{ V _{NaOH} }\] We can simplify n=m/M \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{NaOH} }\]
anonymous
  • anonymous
Now u just plug and chug. V\[V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }\]
anonymous
  • anonymous
Reaction: KHP + NAOH -----> NaKP + H2O
anonymous
  • anonymous
our calculations are based on the dilute concentration of NaOH
Jhannybean
  • Jhannybean
What does the \(\sf C\) stand for?
Jhannybean
  • Jhannybean
Oh, did you mean the Molarity of NaOH represented by C?
anonymous
  • anonymous
yep, molarity and concentration are pretty much the same thing
anonymous
  • anonymous
i think thats all to it
Jhannybean
  • Jhannybean
What i was stuck with was what the heck do I do with the 10 mL of NaOH? Lol
Jhannybean
  • Jhannybean
\[\sf NaOH~(aq)~ + ~KHP ~(aq) ~\rightarrow ~ NaKP ~(aq)~ + H_2O~(l)\] \[\sf M_{NaOH}~ =~ \frac{0.2001~g~KHP ~\times ~ \frac{1~mol~KHP}{204.23~g~KHP}~\times~\frac{1~mol~NaOH}{1~mol~KHP}}{0.01702~L~NaOH}\] but doing it this way made no sense at all.....
anonymous
  • anonymous
yep oops i got it i reckon
Jhannybean
  • Jhannybean
\[\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\]
Jhannybean
  • Jhannybean
Hmm...
Jhannybean
  • Jhannybean
Like what happens to the 6 M NaOH and the 10 mL used to dilute it?is that also extra info given?
anonymous
  • anonymous
we need to find the molarity of the diluted NaOH So \[C _{NaOH}V _{NaOH}=C _{dilute{\space}NaOH}V _{total{\space} volume}\] Where C(NaOH)=6M V(NaOH)=0.01L V(Total)=1.01L Therefore \[C _{dilute {\space} NaOH}=\frac{ 6*0.01 }{ 1.01 }=0.0594M\] Now, molar ratio of NaOH and KHP is 1:1 Therefore, \[n _{NaOH {\space} dilute}=n _{KHP}=\frac{ m _{KHP} }{ M _{KHP} }=\frac{ 0.2001 }{ 204.23 }\] Therefore, \[C _{NaOH}=\frac{ n _{KHP} }{ V _{volume {\space} of {\space} diluted {\space} NaOH} }\] \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{required} }=\frac{ \frac{ 0.2001 }{ 204.23 } }{ 0.01702 }\]
anonymous
  • anonymous
lol wut
anonymous
  • anonymous
that should be right. cause n\[n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{-3}\]
anonymous
  • anonymous
and n\[n _{khp}=\frac{ 0.2001 }{ 204.23 }=9.79 *10^{-4}\]
anonymous
  • anonymous
So \[\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1\] hence molar ratio is 1 so we now that checks
anonymous
  • anonymous
what are your thoughts
aaronq
  • aaronq
the 10 mL 6 M NaOH and 1000 mL of water is extra info (otherwise why not just calculate the molarity from that?), what you guys wrote above is correct: \(\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\)
aaronq
  • aaronq
Is this for a lab, if so do you have to do uncertainty/error propagation? @Jhannybean
Jhannybean
  • Jhannybean
Nope, my lab professor saved us the trouble of doing that :)
Jhannybean
  • Jhannybean
Wow, I didn't expect this problem to be so tricky. I mean I had studied over how to do the calculation, but this question threw me off during the quiz because I did not know how to incorporate the \(\sf V_{diluted ~NaOH}\) and the 6 M of NaOH. If this `wasnt` a 1:1 ratio, how would I have used that information thoug?
aaronq
  • aaronq
like if you had, \(\sf 2~NaOH~(aq)~ + ~H_2X ~(aq) ~\rightarrow ~ NaX ~(aq)~ + 2H_2O~(l)\) \(\sf \dfrac{n_{NaOH}}{2}=\dfrac{n_{H_2X}}{1}\rightarrow n_{NaOH}=2*\dfrac{n_{H_2X}}{1} \) \(\sf C_{NaOH}=\dfrac{n_{NaOH} }{V_{NaOH}}==\dfrac{2*n_{H_2X}}{V_{NaOH}}=\dfrac{2*(\dfrac{mass~of~H_2X}{MW ~of~H_2X})}{V_{NaOH}} \)
anonymous
  • anonymous
@Jhannybean you didn't need a reaction equation to prove the molar ratio was 1:1, my last comments showed how the molar ratio was 1:1 using the first part of the information with the 6Mols etc
anonymous
  • anonymous
and then comparing it to the moles of KHP.

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