Question

25cm3 of a solution of O.2M potassium hydroxide reacts with 30cm3 of a solution of nitric acid. What is the concentration of the acid in g dm-3?

Answer 1

Sorry, I'm late!

Answer 2

what does dm stand for? Decimeter?

Answer 3

Yes.
Zeta, don't be apologetic.

Answer 4

Ok so just use M1V1 = M2V2

25ml NaOH x 0.2M NaOH = M of HNO3 x 30ml HNO3
M of HNO3 = 0.167 M or grams / mol

Answer 5

grams / Liter, not "mol" lol sorry

Answer 6

so....

Answer 7

1L is just 1meter ^3

Answer 8

so...
0.167 mol / 1meter^3 x (10^3 decimeters / 1 meter^3) = 167 mol/decimeter

Answer 9

*correciton again, the initial answer I got was 0.167 mol / L not "g/mol or g/L" LOl sorry

Answer 10

so it should be 167 mol/decimeter^3

Answer 11

My answer is coming to be 10.5.

Answer 12

Let me rewrite this neatly

Answer 13

Ok so just use M1V1 = M2V2

25ml NaOH x 0.2M NaOH = M of HNO3 x 30ml HNO3
M of HNO3 = 0.167 M or 0.167 mol/L

1L = 1 meter^3 so
0.167 mol / 1 meter^3 x ( [10 decimeter]^3 / 1 meter^3) = 167 mol / decimeter^3

Answer 14

oh you want units of "grams" / decimeter ^3 ok sorry lets set that straight

Answer 15

167 mol HNO3 / 1 decimeter^3 x ( 63 grams HNO3 / 1 mol HNO3) = 10521 grams / decimeter^3. I would like it more if the denominator was in meters^3 (or liters) so that the answer would be 10.521 grams / meters^3

Answer 16

So ya you're right the answer is 10.5 but you can only get that if your denominator is "meters^3" NOT "DECImeter^3" as you had told me

Answer 17

so with everything neatly again
Ok so just use M1V1 = M2V2

25ml NaOH x 0.2M NaOH = M of HNO3 x 30ml HNO3
M of HNO3 = 0.167 M or 0.167 mol/L

1L = 1 meter^3 so
0.167 mol / 1 meter^3 x (63 grams HNO3 / 1 mol HNO3) = 10.521 grams / meter^3 of HNO3

Answer 18

First of all, your reaction is the following:\[\rm \color{#C00}1KOH + \color{#C00}1HNO_3 \to \color{#C00}1KNO_3 + \color{#C00}1H_2O\]This tells you that the number of moles of both \(\rm KOH\) and \(\rm HNO_3\) are the same.

The question indirectly gives you the number of moles of \(\rm KOH\). It tells you that the molarity of \(\rm KOH\) is 0.2, meaning that the number of moles of \(\rm KOH\) in \(\rm 1000 cm^3\) of solution is 0.2.

Here, you have \(\rm 25 cm^3\) so the number of moles of \(\rm KOH\) would be \(\frac{25}{1000} = \frac{1}{40}\).

You have the same number of moles for nitric acid (remember why?). So there are \(1/40\) moles of nitric acid which means that its mass is \(1/40 \times (1 + 14 + 48) = 1/40 \times (M_0) = M_0/40 ~\rm grams\).

You are given the volume to be \(\rm 30 cm^3\). \(\rm 1 dm^3\) is \(1000 \rm cm^3\), so the volume of this solution in cubic-decimeters would be \(\frac{30}{1000} =\rm \frac{3}{100} dm^3\).

Dividing the two, you will get\[\rm conc = \dfrac{M_0/40}{3/100} = \dfrac{100 /40 \times M_0}{3} = \dfrac{2.5}{3} \times 63 = 2.5 \times 21 = 52.5\]

Answer 19

Oh no...

Answer 20

Um, no Zeta, that's wrong

Answer 21

I've already solved it

Answer 22

Yes, that's wrong by a factor of five because I did not take the 0.2 into consideration.

The answer would be \(52.5 \times 1/5 = 10.5\)

Answer 23

I think my way is simpler haha

Answer 24

Sure.

Answer 25

I sense some competition here. Anyway, thank you both.

Answer 26

Obviously I know more chem than Zeta haha

Answer 27

We have the same approach though. :)

M1V1 = M2V2 is not necessarily true, so I think we must first list down our reaction and compare like I did.

Answer 28

It's definitely true b/c this is a neutralization reaction so the moles of acid must equal the moles of base. Another way or writing moles is Molarity x Volume (b/c it comes from the equation Molarity = moles / Volume). So moles acid = moles base is the same thing as Macid x Vacid = Mbase x Vbase

Answer 29

But ya you can see the 1:1 mole relationship via the equation too

Answer 30

\[\rm 2NaOH + H_2CO_3 \to N_2CO_3 + 2 NaOH\]

Answer 31

this isn't a neutralization (which occurs between strong acid and strong base) but instead a weak acid - strong base titration

Answer 32

Well, fair enough.

Answer 33

lol

Answer 34

Though looking at it from a stoichiometric point of view, it may not be necessary that the coefficients are always the same. So I guess you're right... darn!!! I give up.

Answer 35

lmao

Answer 36

We're both right, just you did it the long way I did it the short way

Answer 37

Alright I'm out, PEACE!

Answer 38

That's a subjective way to think about it.

I made it one long reply, plus I showed all steps of my work. So yeah.

Answer 39

this is my entire final reply:
25ml NaOH x 0.2M NaOH = M of HNO3 x 30ml HNO3
M of HNO3 = 0.167 M or 0.167 mol/L

1L = 1 meter^3 so
0.167 mol / 1 meter^3 x (63 grams HNO3 / 1 mol HNO3) = 10.521 grams / meter^3 of HNO3

Answer 40

that's it

Answer 41

I'm pretty sure this is objectively shorter than your reply lol

Answer 42

ok.

Answer 43

Sorry im just bored lol

Answer 44

\[\rm moles = \dfrac{25}{1000}\times \dfrac{1}{5} = \dfrac{1}{200}\]\[\rm mass = \dfrac{63}{200} g\]\[\rm volume = \dfrac{30}{1000} dm^3\]\[\rm conc = mass/volume = 10.5 gdm^{-3}\]

Answer 45

You win brah lol. I'm out now. peace