Question

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 \(\sf NaOH\)

Answer 1

@Shalante @Photon336 @aaronq

Answer 2

the \(\sf 1000~ml~H_2O\) is just extra information

Answer 3

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

Answer 4

most likely its 1:1

Answer 5

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
\[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{NaOH} }\]

Answer 6

Now u just plug and chug.

V\[V _{NaOH}=0.01702; m _{KHP}=0.2001g; M _{KHP}=204.23 \frac{ g }{ mol }\]

Answer 7

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

Answer 8

our calculations are based on the dilute concentration of NaOH

Answer 9

What does the \(\sf C\) stand for?

Answer 10

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

Answer 11

yep, molarity and concentration are pretty much the same thing

Answer 12

i think thats all to it

Answer 13

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

Answer 14

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

Answer 15

yep oops i got it i reckon

Answer 16

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

Answer 17

Hmm...

Answer 18

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

Answer 19

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,

\[C _{NaOH}=\frac{ n _{KHP} }{ V _{volume {\space} of {\space} diluted {\space} NaOH} }\]
\[C _{NaOH}=\frac{ \frac{ m _{KHP} }{ M _{KHP} } }{ V _{required} }=\frac{ \frac{ 0.2001 }{ 204.23 } }{ 0.01702 }\]

Answer 20

lol wut

Answer 21

that should be right. cause

n\[n_{dilute {\space} NaOH}=C _{dilute}V _{dilute}=0.0594*0.01702=1.011 *10^{-3}\]

Answer 22

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

Answer 23

So
\[\frac{ n _{NaOH {\space} dilute} }{ n _{KHP} }=0.969...\approx 1\]

hence molar ratio is 1 so we now that checks

Answer 24

what are your thoughts

Answer 25

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:

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

Answer 26

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

Answer 27

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

Answer 28

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?

Answer 29

like if you had, \(\sf 2~NaOH~(aq)~ + ~H_2X ~(aq) ~\rightarrow ~ NaX ~(aq)~ + 2H_2O~(l)\)


\(\sf \dfrac{n_{NaOH}}{2}=\dfrac{n_{H_2X}}{1}\rightarrow n_{NaOH}=2*\dfrac{n_{H_2X}}{1} \)


\(\sf C_{NaOH}=\dfrac{n_{NaOH} }{V_{NaOH}}==\dfrac{2*n_{H_2X}}{V_{NaOH}}=\dfrac{2*(\dfrac{mass~of~H_2X}{MW ~of~H_2X})}{V_{NaOH}} \)

Answer 30

@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

Answer 31

and then comparing it to the moles of KHP.