Question

Help please. An unknown diuretic acid requires 43.86 mL of 0.119 M NaOH to completely neutralize a 0.565 g sample. Calculate the approximate molar mass of the acid.

Answer 1

diuretic?

Answer 2

was it diprotic?

Answer 3

Yes, oops haha

Answer 4

lol so, you know it has 2 acidic protons.

find the moles of NaOH you used, which were twice as many as there was of acid present.

Answer 5

Would that be by multiplying the volume and concentration?

Answer 6

yep

Answer 7

i got .005 moles

Answer 8

though remember that molarity is in liters, so you need to convert mL to L

Answer 9

alright, so now use \(n=\dfrac{m}{M}\) to find the molar mass (M) of the acid

Answer 10

n = moles, m = mass (g) and M = molar mass, correct?

Answer 11

now do I use the mass that's given (.565g)?

Answer 12

yes (to both)

Answer 13

take into account that it's a diprotic acid, though

Answer 14

Sorry, forgot to put the answer choices since i'm getting a different number.
A) 383g/mol
B) 217 g/nol
C) 108g/mol
D) 54g/mol
E) 192g/mol

Answer 15

that fact that it's diuretic, does that mean you double the M when you get it?

Answer 16

diprotic* laptop keeps autocorrecting

Answer 17

no. you use half as many moles of the base

Answer 18

Now i'm confused :(

Answer 19

so when I used that formula i got 113g

Answer 20

okay, suppose \(H_2A\) is a diprotic acid, then:
\(H_2A+ 2NaOH \rightarrow A^- + 2H_2O\)

notice that there are 2 moles of NaOH (from the stoichiometric coefficient)

\(\dfrac{n_{H_2A}}{1}=\dfrac{n_{NaOH}}{2}\)

then supposed you used 1 mole of NaOH

\(\dfrac{n_{H_2A}}{1}=\dfrac{1}{2}\)

\(n_{H_2A}=1/2\)

then you used 1 mole of the acid

Answer 21

half of a mole of the acid**** not 1 mole

Answer 22

SO, however many moles of NaOH you had, divide by 2 and thats how many moles of the acid you had. (ps. make sure you converted mL to L when you found the moles of NaOH)

Answer 23

In that equation your wrote out, where does the Na go?

Answer 24

I'm still not understanding why you needed to half the moles :\

Answer 25

I did get the answer though, which is 217g

Answer 26

sorry, i omitted it because it's not relevant.
you need twice as many OH to neutralize the acid because the acid is releasing 2 protons per molecule of acid (NaOH only releases 1 mole of OH).

Answer 27

a similar (but opposite) scenario would be if you reacted HCl and \(Mg(OH)_2\)

\(HCl +Mg(OH)_2 \rightarrow MgCl_2 +H_2O\)

balancing it:

\(2HCl +Mg(OH)_2 \rightarrow MgCl_2 +2H_2O\)

so you would need TWICE as much acid for a given amount of the base

Answer 28

so you would actually have to balance the equation first, right? How would you know what a diprotic acid looks like written out?

Answer 29

yeah, you would if you didn't know.

common diprotic acids are \(H_2SO_4, H_2CO_3, H_2CrO_4, H_2C_2O_4\)

Answer 30

you would know because they have 2 hydrogens that could (potentially) be donated to a base or solution (depending on strength).

Answer 31

Got it! So, in the first one, since there were more moles of the base (NaOH) you had to half the acid? and in the second one, you had to double it because the balanced equation showed twice as many H+? (I'm trying to put it in simpler terms for myself)

Answer 32

you're a little off still.

each molecule of a monoprotic acid releases 1 acidic hydrogen
each molecule of a diprotic acid releases 2 acidic hydrogens

NaOH only releases 1 OH, so if you neutralized 1 mole of each acid with NaOH, you would need:

1 mole of NaOH for the monoprotic acid and 2 moles of NaOH for the diprotic acid.

you used 2 moles of NaOH for 1 mole of the diprotic acid, you can see the 2:1 ratio.

Answer 33

but if it was a 1:2 ratio I would've had to multiply it with the moles instead of divide?

Answer 34

multiply 2*

Answer 35

well if you were looking for moles of base you would multiply by 2

look at the ratios like this: \(\dfrac{n_{H_2A}}{1}=\dfrac{n_{NaOH}}{2}\)

Answer 36

find how many moles of diprotic acid you need if you use 1 mole of base

Answer 37

1/2

Answer 38

you needed 0.5 moles for 1 mole of base.. does it make sense now?

Answer 39

Yes, sort of. Now would you use this for every questions like this or it depends on the equation?

Answer 40

if they substances don't contribute equivalently, then yes.

Answer 41

great, thank you very much for your help and patience!

Answer 42

no problem dude !