Ask your own question, for FREE!
Chemistry
OpenStudy (anonymous):

50 ML of 0.001 M phosphoric acid are exactly neutralized by the addition of 0.005 N Naoh the molarity of resulting salt is?

OpenStudy (anonymous):

@Callisto help

OpenStudy (callisto):

Sorry.. I haven't learnt it :|

OpenStudy (anonymous):

ok

sam (.sam.):

\[\begin{array}{c} \text{H}_3\text{P}\text{O}_4 \\\end{array}\text{ + }\begin{array}{cc} 3 & \text{Na}\text{O}\text{H} \\\end{array}\text{ $\longrightarrow $ }\begin{array}{cc} 3 & \text{H}_2\text{O} \\\end{array}\text{ + }\begin{array}{c} \text{Na}_3\text{P}\text{O}_4 \\\end{array}\] ------------------------------------ \[M_AV_A=M_BV_B\] \[\text{Molarity of base is}~~ M_B \] \[\text{Molarity of acid is}~~ M_A\] ------------------------------------ \[M_B=\frac{M_AV_A}{V_B}\] Sub in the values

OpenStudy (preetha):

Don forget the stoichiometric ratios. for every mole of phosphoric acid, you need 3 moles of NaOH to neutralize it. So you have to include that factor of 3 in the above calculation. It is a 1: 3 not 1:1 relationship. Try it and see.

OpenStudy (anonymous):

@Kryten help!

OpenStudy (anonymous):

as i see it you got all you need you just need to put numbers in.... only way to learn how to solve this kind of things is to do it yourself...

OpenStudy (anonymous):

i solve it but not getting lol @Kryten thats why i called u

OpenStudy (anonymous):

if u get after solving that then inform me

OpenStudy (anonymous):

The answer should be 6.25 * 10^-4 moles

OpenStudy (anonymous):

but i am not getting that @Kryten

OpenStudy (anonymous):

this is really basic chemistry, begin with writing equation, then write down what you have and what you need and do the magic... put the relationships of reactants and products in equations and just crunch the numbers...

OpenStudy (anonymous):

@.Sam. not getting the answer by ur way

OpenStudy (anonymous):

@Kryten not getting!!!

OpenStudy (anonymous):

i wrote the equation

sam (.sam.):

6.25 * 10^-4 moles This is in moles, I thought you want molarity?

OpenStudy (anonymous):

@Kryten do u have any idea

OpenStudy (anonymous):

i am confused in question it is asked to find molarity but the answer is given in 6.25*10^-4

OpenStudy (anonymous):

yea i have some idea but i gotta run students arent waiting i need to keep them safe... see you later

OpenStudy (anonymous):

@Vincent-Lyon.Fr do u have any idea....

OpenStudy (vincent-lyon.fr):

I won't answer any of your questions as you are pestering this study site without ever doing any effort on your own.

OpenStudy (anonymous):

@Vincent-Lyon.Fr i did but not getting!

OpenStudy (anonymous):

@open_study1 , my answer is same as @Vincent-Lyon.Fr 's

OpenStudy (anonymous):

@shivam_bhalla i tried ny level best but i am unable,

OpenStudy (anonymous):

@.Sam. and @Preetha have given you the valuable hints and tips . Just go ahead and solve it. you have to find the number of moles of H3PO4 initially reacting from given info

OpenStudy (anonymous):

What's M and N?

OpenStudy (anonymous):

M=molarity N=normality

OpenStudy (anonymous):

moles = 5*10^-5 moles

OpenStudy (anonymous):

Yes. Now continue and do the rest of the problem.

OpenStudy (anonymous):

now wat to do

OpenStudy (anonymous):

I'm sorry i don't know this :).

OpenStudy (anonymous):

@shivam_bhalla r u there....

OpenStudy (anonymous):

Plzzz help anyone plzz!

OpenStudy (anonymous):

@Kryten HELP

OpenStudy (anonymous):

I FOUNDED THE MOLE THEN WAT TO DO???

OpenStudy (anonymous):

@Kryten PLZZ GUIDE ME PLZZ

OpenStudy (anonymous):

OH........OH..........PLZZZZ SOME ONE HELP! I AM WAITING FOR 1 DAY PLZZZZ

OpenStudy (anonymous):

5*10^-5 moles

OpenStudy (anonymous):

in your place i would give up on pleading and solve it myself... you have been give instructions by multiple people and you just need to follow them, point is that you want it to be solved so you god forgive wouldnt have to hitch your little grey cells...

OpenStudy (anonymous):

I DID IN THE WAY THEY GUIDED BUT NOT GETTING I FOUNDED THE MOLES THEN WAT TO DO PLZZ TELL ME THE STEPS I WILL SOLVE MY OWN @Kryten

OpenStudy (anonymous):

first send some proof that you did anything (picture of paper of your tries) and then ill help. this is i think your third time you post this question and you just need to read what people have written... look up and read all carefully

OpenStudy (anonymous):

0.15=0.005*V V=30ML

OpenStudy (anonymous):

0.05 MILLI MOLES OF H3PO4 ON REACTING GIVES 0.05 MILLI MOLES OF NA3PO4 IE 5*10^-5 MOLES OF NA3PO4 ARE FORMED

OpenStudy (anonymous):

@Kryten

OpenStudy (anonymous):

DO U NEED MORE INFO?

OpenStudy (anonymous):

now lets start from the begining write first what data you have, chemical reaction equation and then write what you need from data to be calculated

OpenStudy (anonymous):

H3PO4 + 3NaOH ⟶ 3H2O + Na3PO4

OpenStudy (anonymous):

WE HAVE TO FIND THE MOLARITY OF NA3PO4

OpenStudy (anonymous):

write your data properly ex. c(X)= ... V(X)=... c(Y)=.... and then what you need to calculate ex. c(z)=?

OpenStudy (anonymous):

WAT IS C(X)

OpenStudy (anonymous):

that are examples how to write your data properly now lets see you do it

OpenStudy (anonymous):

V(PHOSPHORIC ACID)=50ML M(PHOSPHORIC ACID)=0.001M N(NAOH)=0.005N M(NA3PO4)=?

OpenStudy (anonymous):

IS THAT CORRECT.........

OpenStudy (anonymous):

ok now lets see you have written N(NaOH)=0,005N --> that doesnt exist cause N(NaOH) is marking of number of NaOH atoms and it has no units, and furthermore write it one under another not one next to other cause that is vast. now lets see corection...

OpenStudy (anonymous):

NORMALITY OF NAOH=0.005N

OpenStudy (anonymous):

now please explain to me what is normality and then write all data again properly

OpenStudy (anonymous):

NO OF EQUIVALENT OF SOLUTE IN ONE LITRE OF SOLUTION

OpenStudy (anonymous):

ok now explain me the unit N what is its derivative

OpenStudy (anonymous):

N= WG OF SOLUTE /(EQUIVALENT WG OF SOLUTE * VOLUME(L))

OpenStudy (anonymous):

ok now when you explained it all carry on and write all data properly one item under other and under it all chemical equation

OpenStudy (anonymous):

@Kryten ANYHOW THANZZZ I HAVE TO GO NOW PLZZ GIVE ALL THE HINTS I WILL LOOK IT AFTER WARDS............

OpenStudy (anonymous):

no way, you will write all here so i see what you are doing...

OpenStudy (anonymous):

hello

OpenStudy (anonymous):

@Kryten

OpenStudy (anonymous):

continue where you stoped, write all on one place...

OpenStudy (anonymous):

thats all wat i knw

OpenStudy (anonymous):

but you still havent written all in one place properly

OpenStudy (anonymous):

oh....no wat else u want

OpenStudy (anonymous):

i would like a cold bear and a steak but what you want is the question

OpenStudy (anonymous):

this website was made to help!!!! so plzz

OpenStudy (anonymous):

i am helping you but you are resisting to it... take a look how i solved your past questions where i photographed my work, it was all in sequence like it should be and that is the only way i will help you if you do it properly and as it should be solved and when you get that routine all this will be easy for you so your choice...

OpenStudy (anonymous):

yes u helped me a lot earlier by risiking a lot but i want the reson now why u r not helping that way did any one force u???

OpenStudy (anonymous):

such as @vincent fr @aravind etc......

OpenStudy (anonymous):

no, but i realised that you are taking a path of least resistance which is not good... this site is for learning and not getting all solved with no effort...

OpenStudy (anonymous):

now lets skip all the debate and tell me if you want to do it with my help properly or with no help at all

OpenStudy (anonymous):

i want help

OpenStudy (anonymous):

so whats the problem then... take a look how i solved your past questions and write it all like that here step by step and with understanding and reason why is it like it is for each step so that you LEARN how to solve problems like this...

OpenStudy (anonymous):

ok let me try

OpenStudy (ujjwal):

Its been two days.. LMAO..

OpenStudy (anonymous):

yes no one seems to help me........

OpenStudy (anonymous):

Is the Answer 3.33 x 10^-4 ?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!