Question

What is the molar mass of a monoprotic acid if 27.70 mL of 0.0727 M NaOH is required to neutralize a 0.186-g sample?

Answer 1

Another titration problem?

Answer 2

I think so. We haven't gone over this stuff in class so I'm lost on it pretty much.

Answer 3

I am sorry to hear that

Answer 4

It's fine. None of your doing obviously.

Answer 5

First, \(\large n_{H^+}=n_{OH^-}\). You can easily calculate \(\large n_{OH^-}\) . Since it is a monoprotic acid, \(\large n_{acid}=n_{H^+}\). Finally, you can find MM by divide the mass by moles.

Answer 6

.186g * (1 mol)/ (88.062g) = .0021121483 mol

Answer 7

Where does 88.062g come from?

Answer 8

Is that the MM?

Answer 9

yes

Answer 10

My answer is 92.4g/mol. Can you please show me your work step by step?

Answer 11

What is your answer for moles of \(H^+\)?

Answer 12

I'm lost

Answer 13

Maybe we should slow down a bit. Ok \(\large n_{H^+}=n_{OH^-}=V\times M\)

Answer 14

So (27.70)(.0727) = 2.01379

Answer 15

In this case, V should be in L

Answer 16

.00201379

Answer 17

That's right. Since NaOH and HCl are both monoprotic, \(\large n_{H^+}=n_{OH^-}\); because \(H^++OH^-\rightarrow H_2O\), the ratio is 1:1

Answer 18

Do you get that?

Answer 19

Is it because one is +1 and the other is -1?

Answer 20

Or is that way off?

Answer 21

Yes that's pretty much it

Answer 22

Now that you have moles of \(H^+\), \(MM= \large \frac{mass}{moles}\)

Answer 23

I got 88.06

Answer 24

I did (.186)/(.002112)

Answer 25

Didn't you say .00201379?

Answer 26

I did. I wrote it down wrong on my paper

Answer 27

Just a mistake. What do you have now?

Answer 28

My calculator gave me 92.36

Answer 29

(.186)/(.00201379)

Answer 30

That's my answer too. With 3 sig figs, it should be 92.4g/mol

Answer 31

Right. Good. You're a good tutor

Answer 32

Thanks. If I am good enough, you should be done 30 minutes ago :)

Answer 33

Nah. Chemistry isn't my strong point so it takes me a bit

Answer 34

Quite many homework for the weekend?

Answer 35

Yeah. I'm on 21 out of 25

Answer 36

And I skipped two problem so I'll have to go back to those later

Answer 37

May I help you with the rest?

Answer 38

If you wanna

Answer 39

Great

Answer 40

You want me to post them as a new question so you can get more medals?

Answer 41

Nah. Just do it here

Answer 42

Works for me

Answer 43

A sample of 0.8360 g of an unknown compound containing barium ions (Ba 2+) is dissolved in water and treated with an excess of Na2SO4. If the mass of the BaSO4 precipitate formed is 0.8853 g, what is the percent by mass of Ba in the original unknown compound?

Answer 44

Ok. Let's find \(\% Ba\) in \(BaSO_4\) first.

Answer 45

How you do that?

Answer 46

\(\%Ba=\large \frac{MM_{Ba}}{MM_{BaSO_4}}\)

Answer 47

(137.327)/(233.386) = .5884114728

Answer 48

That's right. Now you can calculate mass of Ba

Answer 49

\(mass~of~Ba=\%~Ba\times mass~of ~BaSO_4\)

Answer 50

Would mass of Ba be 80.80? I took (137.327)(.5884114728)

Answer 51

Oh by "mass", I mean "mass of the BaSO4 precipitate formed is 0.8853 g"

Answer 52

Oh alright

Answer 53

So you would (.8853)(233.386)

Answer 54

=206.6166258

Answer 55

It is (.08853)(.58841)

Answer 56

I see now

Answer 57

Great. It is kind of confusing, right?

Answer 58

.520919373

Answer 59

Yeah it is

Answer 60

How would you know which one to use?

Answer 61

Now come back to our problem: "A sample of 0.8360 g of an unknown compound containing barium ions (Ba 2+) ", "what is the percent by mass of Ba in the original unknown compound?"

Answer 62

You know that mass of Ba = .520919373, you can know calculate percent by mass

Answer 63

RE: "How would you know which one to use?" Like we know what material is needed to answer the question first. Next we will find that material

Answer 64

Ok

Answer 65

For example, to find the answer for this question, I need to know the mass of Ba

Answer 66

So next step: how to find the mass of Ba? How can I find it from the mass of the BaSO4 precipitate formed that is given?

Answer 67

Do you use the MM?

Answer 68

MM of which substance?

Answer 69

Ba since you are looking for it?

Answer 70

Yes, you can use it to find the mass of Ba

Answer 71

We are given mass of \(BaSO_4\) so you can calculate moles of \(BaSO_4\) first. Then you multiply that moles by MM of \(Ba\) to find mass of Ba

Answer 72

There are many ways to solve one problem

Answer 73

You should know how the process goes, like a thinking routine. It will guide you to the right direction, no matter which route you choose

Answer 74

Ok back to our problem, did you find percent by mass of Ba in the unknown substance?

Answer 75

Got .0037932867 moles of BaSO4

Answer 76

Ok, do you know what is the next step?

Answer 77

Timees the moles of BaSO4 by the mass of Ba

Answer 78

*Times

Answer 79

That's right. *the MM, it's different from the mass

Answer 80

.5209206769

Answer 81

That's correct. Now find the percent by mass of Ba in the unknown substance

Answer 82

Would you take what you got above and multiply it by the .8360?

Answer 83

No, to find the percentage, what should we do?

Answer 84

Multiple by 100

Answer 85

That is right. But we always have to divide something by something first, right?

Answer 86

yes

Answer 87

Therefore, you should divide instead of multiply

Answer 88

Would that be where the .8360 comes in?

Answer 89

That's correct

Answer 90

.6231108575

Answer 91

What do you get? Oh, and don't forget...

Answer 92

I got .6231108575

Answer 93

Then after that you times the 100

Answer 94

Right, that's what I mean

Answer 95

62.31108575

Answer 96

Good job

Answer 97

Thank you

Answer 98

Any more problems?