Question

I had this in a lab where HCl+NaOH=H2O+NaCl. The question is at the end of this reaction, how does the (H^+) compare to the (OH)? I don't understand this question, I have the molarity and the volume of the reactants if that helps any. :)

Answer 1

Was this a titration? do you know the pH of the solution at the end?

Answer 2

Yes, it was a titration lab but I do not know the pH the molarity of the acid was .2m and the molarity for the base was .1m.

Answer 3

you mean 0.2 M, right? (make sure you use the right symbol, lowercase "m" means molality)

You should first find the pH of the solution, which is a measure of \([H^+]\)

Once you do that you can compare the \([OH^-]\) and \([H^+]\), because they're related through Kw.

\(K_w=1.0*10^{-14}=[OH^-]*[H^+]\)

I think by compare, here, they mean determine the ratio of them, like "there is 2 times as much of one or the other".

Answer 4

Ok so how would I find the pH of these substances?

Answer 5

what did you start off with?

Answer 6

We didn't deal with the pH but we do have 5.3mL of acid and 10.7mL base the acid was HCl and the base was NaOH.

Answer 7

so you had 5.3mL 0.2 M HCl and 10.7mL 0.1 M NaOH?

Answer 8

Yes :)

Answer 9

okay, so to find the pH, we have to use simple stoichiometry.

Find the moles of each with the molarity equation: \(M=\dfrac{n_{solute}}{L_{solution}}\)

Then cancel them out according to the stoichiometric coefficients, since the ratio is 1:1, there is not much to do. Find the moles.

Answer 10

Ok so what does the n and L stand for?

Answer 11

n = moles, L = liters

Answer 12

Ok got it thank you for the help!!! I 've got to go but I think I can do it from there.

Answer 13

okay, cool.

Answer 14

Will you check my answers please?

Answer 15

@aaronq

Answer 16

sure thing, it's easier to find a mistake (given you made one) if you post your work as well.

Answer 17

I'm sure I made one knowing me and chemistry! haha ok so I converted the ml to liters by dividing the mL values by 1000 since 1000 mL per liters.
Then I took the two liter values and added them together since you said in the equation the liters of the solution.
.0053+.0107=.016 L
Then I took that and plugged it into the formula given along with the molarity values:
.2=n/.016
and
.1=n/.016
then multiply L value to get n by itself for each equation and get Acid: .0032, Base:.0016 I don't know though I think I did it wrong...

Answer 18

hm theres a problem. When you first find the moles, you're not changing the initial volume, that is you dont add up the volumes just yet.

If you do so, it causes the actual number of moles larger than they actually are.

You need to first find the moles of each solution.

Answer 19

Ok I'm sorry I am just so terrible at chemistry and math so stoichiometry is a nightmare for me, I don't even know where to start.

Answer 20

I know, it can be difficult at first. You just need to understand what is going on, unfortunately it makes time and lost of practice lol

So, finding the moles of acid: \(0.2~M=\dfrac{moles}{0.0053~L}\)

moles of acid =0.00106

Find the moles of base

Answer 21

Ok so 0.1=moles/.0107 for base so it's moles= Acid-.00106 and Base- .00107?

Answer 22

so notice that they are essentially the same. If these cancel out in a 1:1 ratio, how many moles do you have left of each after the cancellations?

Answer 23

I don't understand...

Answer 24

if you have 3 friends and each one eats 1 chocolate, and you have 4 chocolates, do you have extra chocolates or hungry friends after you've given all one chocolate?

Answer 25

You have extra chocolate left.

Answer 26

so it's the same concept, each mole of acid wants a mole of base, they cancel each other out. So what do you have left after you've cancelled what you can out?

Answer 27

So I have more base left since .00107 is more than .00106?

Answer 28

how much exactly?

Answer 29

1x10^-5

Answer 30

Thats right. Now that we have the moles of base left, we use the total volume to find the concentration.

This is where you need to add up the individual volumes of each.

Use the molarity formula again to find the concentration of base with the new volume.

Answer 31

I'm lost again why would we use that formula and how since we already know the liters of solution, molarity, and moles right?

Answer 32

what we did first was find the moles of each substance present.

When we mixed the two solutions the acid and base reacted to form water and salt ("they cancelled out"). Since we mixed the solutions, the new volume is larger and because we work in concentrations, the volume is important.

Answer 33

Ok so would I plug in the moles and molarity into the formula to find the new volume?

Answer 34

nope, we plug in the moles and the volume to find the new molarity.

Answer 35

Ok so its .00107/.495for base and .00106/.495 for acid?

Answer 36

hm not quite. remember we "cancelled out" the acid. So all we have left is base, \(1*10^{-5}\) moles of base

Answer 37

Oh so is it Base: 1.10^-5/.495?

Answer 38

You have the right idea, but your numbers are wrong.

the new volume is 5.3mL+ 10.7mL= 16 mL or 0.016 L

so its \(M=\dfrac{1.0*10^{-5}}{0.016~L}=0.000625 =6.25*10^{-4}~M\)

Answer 39

Ok I see now I just did the adding wrong!

Answer 40

yep! and to finish the question:

This is base right? so it's \([OH^-]=6.25∗10^{−4}~M \)

Answer 41

use the \(K_w=1.0*10^{-14}=[OH^-]*[H^+]\) to find \([H^+]\)

Answer 42

Ok thank you!!

Answer 43

no problem! ask if you any more doubts

Answer 44

Is it 1.0x10^-14/6.25x10^-4=1.6x10^-19 by any chance?

Answer 45

@aaronq

Answer 46

it's \(1.6*10^{-11}\)

Answer 47

Ok thanks again! Sorry for the slow figuring out of things on my part haha

Answer 48

no worries! no one is born a genius - at least you're working towards an understanding which will ultimately benefit you in one way or another.

Answer 49

I hope so haha hopefully in the near future too!

Answer 50

haha just keep at it!