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Jhannybean

  • one year ago

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 \(\sf NaOH\)

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  1. Jhannybean
    • one year ago
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    @Shalante @Photon336 @aaronq

  2. Jhannybean
    • one year ago
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    the \(\sf 1000~ml~H_2O\) is just extra information

  3. anonymous
    • one year ago
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    you need to know the molar ratio's of NaOH and KHP

  4. anonymous
    • one year ago
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    most likely its 1:1

  5. anonymous
    • one year ago
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    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 \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{NaOH} }\]

  6. anonymous
    • one year ago
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    Now u just plug and chug. V\[V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }\]

  7. anonymous
    • one year ago
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    Reaction: KHP + NAOH -----> NaKP + H2O

  8. anonymous
    • one year ago
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    our calculations are based on the dilute concentration of NaOH

  9. Jhannybean
    • one year ago
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    What does the \(\sf C\) stand for?

  10. Jhannybean
    • one year ago
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    Oh, did you mean the Molarity of NaOH represented by C?

  11. anonymous
    • one year ago
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    yep, molarity and concentration are pretty much the same thing

  12. anonymous
    • one year ago
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    i think thats all to it

  13. Jhannybean
    • one year ago
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    What i was stuck with was what the heck do I do with the 10 mL of NaOH? Lol

  14. Jhannybean
    • one year ago
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    \[\sf NaOH~(aq)~ + ~KHP ~(aq) ~\rightarrow ~ NaKP ~(aq)~ + H_2O~(l)\] \[\sf M_{NaOH}~ =~ \frac{0.2001~g~KHP ~\times ~ \frac{1~mol~KHP}{204.23~g~KHP}~\times~\frac{1~mol~NaOH}{1~mol~KHP}}{0.01702~L~NaOH}\] but doing it this way made no sense at all.....

  15. anonymous
    • one year ago
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    yep oops i got it i reckon

  16. Jhannybean
    • one year ago
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    \[\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\]

  17. Jhannybean
    • one year ago
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    Hmm...

  18. Jhannybean
    • one year ago
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    Like what happens to the 6 M NaOH and the 10 mL used to dilute it?is that also extra info given?

  19. anonymous
    • one year ago
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    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, \[C _{NaOH}=\frac{ n _{KHP} }{ V _{volume {\space} of {\space} diluted {\space} NaOH} }\] \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{required} }=\frac{ \frac{ 0.2001 }{ 204.23 } }{ 0.01702 }\]

  20. anonymous
    • one year ago
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    lol wut

  21. anonymous
    • one year ago
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    that should be right. cause n\[n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{-3}\]

  22. anonymous
    • one year ago
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    and n\[n _{khp}=\frac{ 0.2001 }{ 204.23 }=9.79 *10^{-4}\]

  23. anonymous
    • one year ago
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    So \[\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1\] hence molar ratio is 1 so we now that checks

  24. anonymous
    • one year ago
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    what are your thoughts

  25. aaronq
    • one year ago
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    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: \(\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\)

  26. aaronq
    • one year ago
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    Is this for a lab, if so do you have to do uncertainty/error propagation? @Jhannybean

  27. Jhannybean
    • one year ago
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    Nope, my lab professor saved us the trouble of doing that :)

  28. Jhannybean
    • one year ago
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    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?

  29. aaronq
    • one year ago
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    like if you had, \(\sf 2~NaOH~(aq)~ + ~H_2X ~(aq) ~\rightarrow ~ NaX ~(aq)~ + 2H_2O~(l)\) \(\sf \dfrac{n_{NaOH}}{2}=\dfrac{n_{H_2X}}{1}\rightarrow n_{NaOH}=2*\dfrac{n_{H_2X}}{1} \) \(\sf C_{NaOH}=\dfrac{n_{NaOH} }{V_{NaOH}}==\dfrac{2*n_{H_2X}}{V_{NaOH}}=\dfrac{2*(\dfrac{mass~of~H_2X}{MW ~of~H_2X})}{V_{NaOH}} \)

  30. anonymous
    • one year ago
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    @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

  31. anonymous
    • one year ago
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    and then comparing it to the moles of KHP.

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