## 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 $$\sf NaOH$$

1. Jhannybean

@Shalante @Photon336 @aaronq

2. Jhannybean

the $$\sf 1000~ml~H_2O$$ is just extra information

3. anonymous

you need to know the molar ratio's of NaOH and KHP

4. anonymous

most likely its 1:1

5. 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 $C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{NaOH} }$

6. anonymous

Now u just plug and chug. V$V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }$

7. anonymous

Reaction: KHP + NAOH -----> NaKP + H2O

8. anonymous

our calculations are based on the dilute concentration of NaOH

9. Jhannybean

What does the $$\sf C$$ stand for?

10. Jhannybean

Oh, did you mean the Molarity of NaOH represented by C?

11. anonymous

yep, molarity and concentration are pretty much the same thing

12. anonymous

i think thats all to it

13. Jhannybean

What i was stuck with was what the heck do I do with the 10 mL of NaOH? Lol

14. 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.....

15. anonymous

yep oops i got it i reckon

16. Jhannybean

$\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}$

17. Jhannybean

Hmm...

18. Jhannybean

Like what happens to the 6 M NaOH and the 10 mL used to dilute it?is that also extra info given?

19. 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, $C _{NaOH}=\frac{ n _{KHP} }{ V _{volume {\space} of {\space} diluted {\space} NaOH} }$ $C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{required} }=\frac{ \frac{ 0.2001 }{ 204.23 } }{ 0.01702 }$

20. anonymous

lol wut

21. anonymous

that should be right. cause n$n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{-3}$

22. anonymous

and n$n _{khp}=\frac{ 0.2001 }{ 204.23 }=9.79 *10^{-4}$

23. anonymous

So $\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1$ hence molar ratio is 1 so we now that checks

24. anonymous

25. 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: $$\sf C_{NaOH}=\dfrac{\dfrac{0.2001~g~KHP}{204.23~g/mol~KHP}}{0.01702~L~NaOH} = 0.05757~M_{NaOH}$$

26. aaronq

Is this for a lab, if so do you have to do uncertainty/error propagation? @Jhannybean

27. Jhannybean

Nope, my lab professor saved us the trouble of doing that :)

28. 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?

29. aaronq

like if you had, $$\sf 2~NaOH~(aq)~ + ~H_2X ~(aq) ~\rightarrow ~ NaX ~(aq)~ + 2H_2O~(l)$$ $$\sf \dfrac{n_{NaOH}}{2}=\dfrac{n_{H_2X}}{1}\rightarrow n_{NaOH}=2*\dfrac{n_{H_2X}}{1}$$ $$\sf C_{NaOH}=\dfrac{n_{NaOH} }{V_{NaOH}}==\dfrac{2*n_{H_2X}}{V_{NaOH}}=\dfrac{2*(\dfrac{mass~of~H_2X}{MW ~of~H_2X})}{V_{NaOH}}$$

30. 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

31. anonymous

and then comparing it to the moles of KHP.