Normality vs molarity

[spoog74]

just trying to make sure i understand this FULLY.


Determine the normality and molarity of a 100 mL H2SO4 solution that requires 75mL of .5000 N KOH for complete neutralization..


The answer is .3750 N and .1875 M

My question is ;

For normality, dont you multiply the N by 2? So wouldnt the N be the .1875 and the M be the .3750?

I guess im confused and would LOVE some input. THank you all

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[AllahIsGreater]

N is always greater than or equal to Molarity. You have the concepts, right you just switched them around!

 

[lyjw88]

so in neutralization, the multiplying factor (to convert m to N) is determined by the # of ionizable H+/OH- ions right?

 

[AllahIsGreater]

so in neutralization, the multiplying factor (to convert m to N) is determined by the # of ionizable H+/OH- ions right?

exactly!

 

[bharat008]

QUICK & SIMPLE EXPLANATION:

Normality is always a whole-number multiple of molarity.
Normality = equiv # x Molarity
In acid-base reactions, the equiv # is the number of H+ or OH- available in a formula unit of the substance. In ox-red reactions, the equiv # is the number of electrons gained or lost by one formula unit of the substance.

MORE IN-DEPTH EXPLANATION:

Molarity = moles/Liter
Normality = equivalents solute / liters sol'n

To get the equivalent #
Acid-base reactions the equiv. is the quant. that supplies 1 mol of H+ or 1 mol of OH-
EX:
HCL yields 1 H+ (equiv # = 1)
H2SO4 yields 2 H+ (equiv # = 2)
H3PO4 yields 3 H+ (equiv # = 3)
Al(OH)3 yields 3 OH- (equiv # = 3)

In an oxidation-reduction, an equiv. is the quant. of substance that gains or loses 1 mol of electrons.
EX:
KMnO4 when produces Mn2+, (red of 5 electrons (equiv#=5)
KMnO4 when produces MnO2, (red of 3 electrons (equiv #=3)
Na2C2O4 when produces CO2, (oxid of 2 electrons (equiv #=2)

For example, when H2SO4 reacts as an acid to form SO4-2, it loses two H+ ions. So, 1 mol of H2SO4 will be 2 equivalents. If 1 mol of H2SO4 is dissolved in enough water to form 1 L of solution, its conc can be expressed as either 1M or 2 N:
1 mol/1L = 1M
OR 2 equivalents / 1L = 2N

Hope this helps. I didn't write this explanation but whoever did it's really good. I just learned this concept since I had no clue what the hell normality is.. 👍

 

[mmoosavi41]

okay so to clarify:

to find the molarity of H2SO4 in the example spoog gave you do M1V1=M2V2, right?

but how do you find the normality? there are two equivalents of H+ in H2SO4 so do you do 2/.1 L and for KOH you do 1/.075? is that right or am i totally wrong? and once u find the normality of KOH and H2SO4, do you just add them up to get the overall normality?

 

[bharat008]

okay so to clarify:

to find the molarity of H2SO4 in the example spoog gave you do M1V1=M2V2, right?

but how do you find the normality? there are two equivalents of H+ in H2SO4 so do you do 2/.1 L and for KOH you do 1/.075? is that right or am i totally wrong? and once u find the normality of KOH and H2SO4, do you just add them up to get the overall normality?

First, find the molarity as you always do using NMV = NMV.. This will give you molarity. Then to find the Normality, the # of ionizable protons is 2 (due to H2SO4) so it becomes MOLARITY*2 = Normality..

Always multiply molarity times the ionizable protons to get NOrmality..

Hope that makes sense..

 

[rah08e]

I actually found the easiest way to calculate this problem is to use the formula:

1.VaNa = VbNb (Where V= volume and N= Normality

2.Convert volume to Liters

3.Since H2SO4 is Diprotic (will donate 2 H+), the Molarity is MULTIPLIED by 2
4.Since KOH is Monoprotic (will donate 1 H+) the Normality will be the SAME as the Molarity

5. (.1)(2M) = (.075)(.5) ---> M= .1875

- Only tricky part here is the "2M" I used instead of "Na." Since we do not know Molarity of H2SO4, but we know that it is diprotic, we can represent Na (the Normality of "a") as 2M (2 times the Molarity). If the unknown Molarity was monoprotic (like KOH or acetic acid), Na would EQUAL Molarity

6. Now take M (.1875) and multiply it by 2.... you know the drill

let me know if this was helpful

-Ross, FSU Pre Dental

 

[spoog74]

QUICK & SIMPLE EXPLANATION:

Normality is always a whole-number multiple of molarity.
Normality = equiv # x Molarity
In acid-base reactions, the equiv # is the number of H+ or OH- available in a formula unit of the substance. In ox-red reactions, the equiv # is the number of electrons gained or lost by one formula unit of the substance.

MORE IN-DEPTH EXPLANATION:

Molarity = moles/Liter
Normality = equivalents solute / liters sol'n

To get the equivalent #
Acid-base reactions the equiv. is the quant. that supplies 1 mol of H+ or 1 mol of OH-
EX:
HCL yields 1 H+ (equiv # = 1)
H2SO4 yields 2 H+ (equiv # = 2)
H3PO4 yields 3 H+ (equiv # = 3)
Al(OH)3 yields 3 OH- (equiv # = 3)

In an oxidation-reduction, an equiv. is the quant. of substance that gains or loses 1 mol of electrons.
EX:
KMnO4 when produces Mn2+, (red of 5 electrons (equiv#=5)
KMnO4 when produces MnO2, (red of 3 electrons (equiv #=3)
Na2C2O4 when produces CO2, (oxid of 2 electrons (equiv #=2)

For example, when H2SO4 reacts as an acid to form SO4-2, it loses two H+ ions. So, 1 mol of H2SO4 will be 2 equivalents. If 1 mol of H2SO4 is dissolved in enough water to form 1 L of solution, its conc can be expressed as either 1M or 2 N:
1 mol/1L = 1M
OR 2 equivalents / 1L = 2N

Hope this helps. I didn't write this explanation but whoever did it's really good. I just learned this concept since I had no clue what the hell normality is.. 👍

this was great, thank you all so much for the help. Gd bless

 

[MikeQ]

Did anybody get this question on the DAT?

If they were to ask you for molarity, and you are dealing with a base which dissociates twice. Would you take it into consideration if asked for molarity or would you ignore it?

Meaning: HCl + Ba(OH)2 - will you set up your equation: n1M1V1 = n2M2V2 - n being for moles. I know essentially this is normality so if asked for molarity, will you ignore the n.

Thank you!

 

[hope_to_match]

Did anybody get this question on the DAT?

If they were to ask you for molarity, and you are dealing with a base which dissociates twice. Would you take it into consideration if asked for molarity or would you ignore it?

Meaning: HCl + Ba(OH)2 - will you set up your equation: n1M1V1 = n2M2V2 - n being for moles. I know essentially this is normality so if asked for molarity, will you ignore the n.

Thank you!

if you get a molecule that has 2 OH's for example, just multiply the molarity by 2

 

[MikeQ]

if you get a molecule that has 2 OH's for example, just multiply the molarity by 2

Thank you for the reply. I am aware of this but then that would be solving for normality when you take into account the two OH.

 

[hope_to_match]

Thank you for the reply. I am aware of this but then that would be solving for normality when you take into account the two OH.

isnt that what you were asking? im sorry i didn;t get a chance to read all replies