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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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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the \(\sf 1000~ml~H_2O\) is just extra information
you need to know the molar ratio's of NaOH and KHP

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Other answers:

most likely its 1:1
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 \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{NaOH} }\]
Now u just plug and chug. V\[V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }\]
Reaction: KHP + NAOH -----> NaKP + H2O
our calculations are based on the dilute concentration of NaOH
What does the \(\sf C\) stand for?
Oh, did you mean the Molarity of NaOH represented by C?
yep, molarity and concentration are pretty much the same thing
i think thats all to it
What i was stuck with was what the heck do I do with the 10 mL of NaOH? Lol
\[\sf NaOH~(aq)~ + ~KHP ~(aq) ~\rightarrow ~ NaKP ~(aq)~ + H_2O~(l)\] \[\sf M_{NaOH}~ =~ \frac{0.2001~g~KHP ~\times ~ \frac{1~mol~KHP}{204.23~g~KHP}~\times~\frac{1~mol~NaOH}{1~mol~KHP}}{0.01702~L~NaOH}\] but doing it this way made no sense at all.....
yep oops i got it i reckon
\[\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\]
Hmm...
Like what happens to the 6 M NaOH and the 10 mL used to dilute it?is that also extra info given?
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, \[C _{NaOH}=\frac{ n _{KHP} }{ V _{volume {\space} of {\space} diluted {\space} NaOH} }\] \[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{required} }=\frac{ \frac{ 0.2001 }{ 204.23 } }{ 0.01702 }\]
lol wut
that should be right. cause n\[n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{-3}\]
and n\[n _{khp}=\frac{ 0.2001 }{ 204.23 }=9.79 *10^{-4}\]
So \[\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1\] hence molar ratio is 1 so we now that checks
what are your thoughts
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: \(\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\)
Is this for a lab, if so do you have to do uncertainty/error propagation? @Jhannybean
Nope, my lab professor saved us the trouble of doing that :)
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?
like if you had, \(\sf 2~NaOH~(aq)~ + ~H_2X ~(aq) ~\rightarrow ~ NaX ~(aq)~ + 2H_2O~(l)\) \(\sf \dfrac{n_{NaOH}}{2}=\dfrac{n_{H_2X}}{1}\rightarrow n_{NaOH}=2*\dfrac{n_{H_2X}}{1} \) \(\sf C_{NaOH}=\dfrac{n_{NaOH} }{V_{NaOH}}==\dfrac{2*n_{H_2X}}{V_{NaOH}}=\dfrac{2*(\dfrac{mass~of~H_2X}{MW ~of~H_2X})}{V_{NaOH}} \)
@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
and then comparing it to the moles of KHP.

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