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CHEM HELP

Blogs > miNi
Post a Reply
miNi
Profile Blog Joined October 2005
Korea (South)2010 Posts
February 25 2008 05:09 GMT
#1
k, it's my turn now. Here are some pre-lab questions which i have to complete, but i dont know how to do.

1. calculate the molarity of a solution of sodium hydroxide if 23.64mL of this solution is needed to neutralize .5632g of potassium hydrogen phthalate.

2. It is found that 24.68mL of .1165M NaOH is needed to titrate .2931 g of an unknown acid to the phenolphthalein end point. Calculate the equivalent mass of the acid.

help appreciated

GrayArea
Profile Blog Joined December 2007
United States872 Posts
February 25 2008 05:13 GMT
#2
First one I think is simply using the molarity equation. Molarity = moles/liters. So I would convert the grams of solid to moles and divide by the volume of liquid to get the molarity.

Second one I am not really sure.
Kang Min Fighting!
miNi
Profile Blog Joined October 2005
Korea (South)2010 Posts
February 25 2008 05:16 GMT
#3
i'm sure it's a little more complex than that
it's an acid-base neutralization exercise
GrayArea
Profile Blog Joined December 2007
United States872 Posts
February 25 2008 05:20 GMT
#4
what chemistry is this for? high school or college?
Kang Min Fighting!
miNi
Profile Blog Joined October 2005
Korea (South)2010 Posts
February 25 2008 05:22 GMT
#5
ap chem
SIUnit
Profile Joined December 2006
China288 Posts
February 25 2008 05:25 GMT
#6
For first one I think balance out the equations, then find how much H+ is released by the potassium hydrogen phthalate, and calculate the moles of OH- used, divide by volume for molarity.
SayaSP
Profile Blog Joined February 2007
Laos5494 Posts
February 25 2008 05:26 GMT
#7
I just did a whole bunch of stuff on molarity molality stuff but im just dumb
[iHs]SSP | I-NO-KI BOM-BA-YE | のヮの http://tinyurl.com/MLIStheCV , MLIS.
miNi
Profile Blog Joined October 2005
Korea (South)2010 Posts
February 25 2008 05:28 GMT
#8
= 2.758e-3 mol KHP,

cant find out the [H+] because KHP is a weak acid
SIUnit
Profile Joined December 2006
China288 Posts
February 25 2008 05:31 GMT
#9
Also, for the second one, do you know if the acid is monoprotic or something else? Equivalent mass is molar mass divided by number of H+ released by the compound. It doesn't seem like you can calculate the equivalent mass if you don't know how many H+ the compound releases.
SayaSP
Profile Blog Joined February 2007
Laos5494 Posts
February 25 2008 05:31 GMT
#10
Ok, for #1 I think that in order for a base to neutralize and acid it the moles of the two are equal, correct? So I think you would find the moles of KHP, which is blah/204.22 and since the moles would be equal you have the information to calculate molarity. Someone correct me please hhaha
[iHs]SSP | I-NO-KI BOM-BA-YE | のヮの http://tinyurl.com/MLIStheCV , MLIS.
GrayArea
Profile Blog Joined December 2007
United States872 Posts
February 25 2008 05:32 GMT
#11
dude, im taking organic chemistry in college, i should know this stuff but i dont. -_- i feel like such a failure right now..sigh..

in my defense, i took general chem 2 years ago, so ya..but still, damn...sry man
Kang Min Fighting!
SIUnit
Profile Joined December 2006
China288 Posts
Last Edited: 2008-02-25 05:40:28
February 25 2008 05:37 GMT
#12
Haha Gray, same here. 2nd semester orgo and getting stumped by gen chem...

Ok, for the first question, ignore my first response. At the endpoint, the number of moles of NaOH used in the titration is equal to the number of moles of KHP (Molar mass 204.2 g/mole) weighed out. Knowing the volume of base and number of moles of acid, you can calculate the concentration of the NaOH solution using the following expression

Macid * Vacid = Mbase * Vbase
Macid = molarity (moles/L) of acid
Vacid = volume (L) of acid
Mbase = molarity (moles/L) of base
Vbase = volume (L) of base

Which is basically the moles of KHP over liters of NaOH used.

While KHP is a weak acid, it does not change the fact that when the end point is reached, theoretically the moles of OH- should equal the moles of H+ released from the acid.
SayaSP
Profile Blog Joined February 2007
Laos5494 Posts
February 25 2008 05:40 GMT
#13
Hey thats what I said! :D *feels smart* lol
[iHs]SSP | I-NO-KI BOM-BA-YE | のヮの http://tinyurl.com/MLIStheCV , MLIS.
Steelflight-Rx
Profile Blog Joined July 2007
United States1389 Posts
February 25 2008 06:01 GMT
#14
On February 25 2008 14:32 GrayArea wrote:
dude, im taking organic chemistry in college, i should know this stuff but i dont. -_-


rofl same here, i was just thinking that to myself

yubee wrote: you know? it's a great night you should all smile no matter what harddships, because grass grows and the sky is blue and it's a good life.
xhuwin
Profile Blog Joined June 2007
United States476 Posts
February 25 2008 06:03 GMT
#15
#2

Well I'm gonna assume the acid is monoprotic. Anyway this is just stoichiometry:

Moles Base = Moles Acid at the titration end point.
Moles Base = .02468 L * .1165 M NaOH = .2931 g Acid / (X g per mol Acid) ---- x is what you're solving for, the molar mass of the acid.

Basically, for the base, we multiply volume by concentration, and for acid, we divide mass by molar mass. Both sides of the equation give us a number of moles, which we know is equal because we're talking about the titration end point (assuming a monoprotic acid).

PM me if you still need help!
xyn
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
February 25 2008 06:14 GMT
#16
On February 25 2008 14:37 SIUnit wrote:
Macid * Vacid = Mbase * Vbase

This should be the only equation you need I'm pretty sure.
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