A community for students.
Here's the question you clicked on:
 0 viewing
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 \(\sf NaOH\)
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 \(\sf NaOH\)

This Question is Closed

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0@Shalante @Photon336 @aaronq

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0the \(\sf 1000~ml~H_2O\) is just extra information

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need to know the molar ratio's of NaOH and KHP

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0Now u just plug and chug. V\[V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Reaction: KHP + NAOH > NaKP + H2O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0our calculations are based on the dilute concentration of NaOH

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0What does the \(\sf C\) stand for?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Oh, did you mean the Molarity of NaOH represented by C?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, molarity and concentration are pretty much the same thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think thats all to it

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0What i was stuck with was what the heck do I do with the 10 mL of NaOH? Lol

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf NaOH~(aq)~ + ~KHP ~(aq) ~\rightarrow ~ NaKP ~(aq)~ + H_2O~(l)\] \[\sf M_{NaOH}~ =~ \frac{0.2001~g~KHP ~\times ~ \frac{1~mol~KHP}{204.23~g~KHP}~\times~\frac{1~mol~NaOH}{1~mol~KHP}}{0.01702~L~NaOH}\] but doing it this way made no sense at all.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep oops i got it i reckon

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Like what happens to the 6 M NaOH and the 10 mL used to dilute it?is that also extra info given?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0that should be right. cause n\[n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and n\[n _{khp}=\frac{ 0.2001 }{ 204.23 }=9.79 *10^{4}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \[\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1\] hence molar ratio is 1 so we now that checks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are your thoughts

aaronq
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0Is this for a lab, if so do you have to do uncertainty/error propagation? @Jhannybean

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Nope, my lab professor saved us the trouble of doing that :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Wow, 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
 one year ago
Best ResponseYou've already chosen the best response.0like 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0and then comparing it to the moles of KHP.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.