Question

1. Determine the volume of sodium hydroxide added to the flask from the buret.

2. Calculate the molarity of the Hydrochloric acid in the flask.

Information:
Initial volume of NaOH in buret: 0 mL
Volume of HCI in flask: 20 mL
pH when 0 mL of NaOH has been added: 0.70
Volume of NaOH remaining at end point: 30 mL
pH each time 5 mL of NaOH is added: (amount in mL|pH:

5 mL|0.96 pH
10 mL|1.3 pH
15 mL|2.15 pH
20 mL|7.00 pH (now neutral)
25 mL|12.7 pH
30 mL|12.85 pH
35 mL|12.94 pH
40 mL|13 pH
45 mL|13.05 pH
50 mL|13.08 pH

Thanks so much!!

Answer 1

@nincompoop @ikram002p @oldrin.bataku @TheAsker2002 @BloomLocke367 @dmndlife24 @rvc @JoannaBlackwelder

Answer 2

I am guessing what it is looking for in the first question is, how much sodium hydroxide is added to the flask to get to the equivalence point (which for strong acids mixed with strong bases is a pH of 7)

Answer 3

Does that make sense, @lovelylolita55 ?

Answer 4

Yes, you are correct @JoannaBlackwelder

Answer 5

Sorry I took forever to reply @JoannaBlackwelder

Answer 6

No worries. So, how much volume is NaOH is that?

Answer 7

30 mL?

Answer 8

Hm, it looks like to me the data shows it is a pH of 7 at 20 mL added.

Answer 9

Oh that's what you meant. Yes, pH of 7 at 20 mL.

Answer 10

:-) And for the next part, do you have the molarity of NaOH and the balanced reaction?

Answer 11

Yes, the buret was filled with 0.25 molar NaOH solution.

Answer 12

Nice, and the reaction?

Answer 13

I'm not sure I have that info..This was a strong acid - strong base neutralization reaction between NaOH and HCI

Answer 14

Is that what you're asking for? :P

Answer 15

Yep, do you know the general form of a neutralization reaction?

Answer 16

No, I don't think I do

Answer 17

All strong acid/strong base reactions form water and a salt

Answer 18

Oh I see

Answer 19

So the molarity of the HCI would be? I don't know how to calculate that

Answer 20

Well, when we write the reaction:
HCl+NaOH->H2O+NaCl

Answer 21

We have a balanced reaction already, since the number of each element is the same on each side.

Answer 22

So that means we have a 1:1 molar ratio of HCl and NaOH

Answer 23

Ok

Answer 24

We can then use the molarity of the NaOH and the volume to find how many moles of NaOH were needed in the reaction.

Answer 25

Can you do that step?

Answer 26

So concentration and volume?

Answer 27

Yep

Answer 28

So the volume of HCI would be 20 mL, but I'm not sure about the molarity.

Answer 29

We have both for NaOH though.

Answer 30

We are looking for the molarity of HCl

Answer 31

Oh I see, so they would be equal? Since the ratio is 1:1

Answer 32

The number of moles are equal.

Answer 33

ok

Answer 34

Hmm I'm having a bit of trouble remembering how to do this :'(

Answer 35

So, the molarity formula is
Molarity=mols solute/L solution

Answer 36

Can you use that to calculate moles of NaOH?

Answer 37

0.0125...is that terribly wrong? lol

Answer 38

Hm, how did you get that?

Answer 39

M = 0.25 mols / 20 mL...I messed that up, didn't I?

Answer 40

Yeah, the .25 isn't the moles it is the M (molarity)

Answer 41

And the molarity formula requires the volume to be in L, so we need to convert mL->L

Answer 42

Oh i see

Answer 43

:-)

Answer 44

0.02 L then?

Answer 45

Yep

Answer 46

I'm not sure what the amount of moles are then

Answer 47

Well, the formula with what we know plugged in is:
.25=moles NaOH/.02

Answer 48

Do you see how I substituted?

Answer 49

Yes, that's brilliant lol

Answer 50

:D haha, thanks! Can you solve for moles?

Answer 51

I just got another out of wack number..could you maybe walk me through that one too? You're helping me out so much!

Answer 52

What number did you get? And I'm glad i can help. :-)

Answer 53

12.5..but that didn't seem right?

Answer 54

Yeah, I did it this way: where x stands for moles of NaOH