Question

Medal and fan

A weak acid with an intitial pH of 3.2 was titrated with a strong base. 15 mL of .1 M NaOH was added to the acid to reacht eh equivalence point at a pH of 8.6. What would you expect the approximate pH of the analyte to be after the first 5 mL of .1 M NaOH was added?

I know the general equations to use, but I'm not exactly sure how to solve it. If you could break it down and show me how to solve it, that would be awesome, thankyou(:

Answer 1

@Somy Would you know how to solve this?

Answer 2

hmm im trying to find relation here from what's written in my book
but can't seem to find specific way of solving this

Answer 3

what equations u used??

Answer 4

An acid can't have a "pH". It can only have a pKa. If you mean an acidic "solution" has a pH, then that would make sense but I"m not sure.

Answer 5

Um...an acid can have a pH lol. pKa is the dissociation constant but in a log measure.

Answer 6

No, pH = -log[H3O+}, thus it's a measure of how much hydronium cations (or protons) are in a solution DUE to an acid. The amount of protons in solution depends on the pKa of an acid

Answer 7

The pH of the solution is different than the pH of the acid....

Answer 8

@Abhisar

Answer 9

So like I said, an acidic compound doesn't have it's own pH but instead, it has a pKa.

Answer 10

@aprehan @seehearcreate
\(\bigstar\) pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale

Answer 11

Yes, I know that Abhisar. pH = -log[H+] thus pH is inversely related to the H+ concentration in solution

Answer 12

Alright, that isn't what I'm asking though. The question is the question, it's not worded in my own words, but in my class. That's what I need help with.

Answer 13

It's worded incorrectly, but I'll assume 3.2 refers to the pH of "solution"

Answer 14

As I said, not in my own wording. It's the class. Online is flawed in many ways.

Answer 15

but in general if someone says that pH of an acid is x.x, then it's understood that the acid is in solution and it's pH is x.x

Answer 16

Right. I gotcha.

Answer 17

ok let's see the origial question now

Answer 18

*original

Answer 19

Ok so I'm trying to solve this and ima going to talk as i go. We're indirectly given the pKa of the acid b/c we're told that the eq pt has a pH of 8.6 so the pKa of the acid is also 8.6

Answer 20

I know this b/c at the eq point, the concentration of the acid and its conj base are equal to each other so the ratio "[Conj base]/[acid]" is equal to 1 so the H-H equation will go from pH = pKa + log [conj base]/[acid] towards pH = pKa

Answer 21

the log[conj base]/[acid] term dissapears b/c the ratio of [conj base]/[acid] is 1 and the log of 1 is - 0

Answer 22

so what can i do next....

Answer 23

I don't understand how you acquired the concentration of the acid...

Answer 24

No, the pKa of 8.6 of the acid is not a concentration -_-

Answer 25

See, at the equivalence point, you have equal amounts of acid and its conjugate base right?

Answer 26

I didn't say that.

Answer 27

Your question says the pH at the equivalence point is 8.6. I'm just defining to you what the equivalence point means

Answer 28

To acquire the log portion, you need the base over the acid, the concentration of the base being .1 but the concentration of the acid was not stated.

Answer 29

yes you're right about the "initial" concentration of the acid not being stated. That's something I'm trying to figure out. But b/c the question mentions the phrase "equivalence point", you know that when the equivalence point is reached, the concentration of the acid and base are equal to each other

Answer 30

By the way.... pKa is only equal to the pH at the half equivalence point...

Answer 31

yes exactly

Answer 32

oh wait

Answer 33

LOL you're totally right, i'm tripping

Answer 34

Yeah haha.

Answer 35

Okay, at the equivalence point, all of the acid has been neutralized into it's conjugate base. So we only have conjugate base at the equivalence point and noething else (besides water)

Answer 36

agree?

Answer 37

and this makes sense b/c the pH at the equivalence point is greater than 7 (thus basic) b/c of the presence of only conjugate base

Answer 38

Yes, because at EP, the OH- ions and H+ equal each other.

Answer 39

ya okay good

Answer 40

Ok im going to pull out my acid/base notes from 2 years ago lol

Answer 41

just for you

Answer 42

Haha, thanks(:

Answer 43

Ok so we have 15ml of 0.1M NaOH. We can find the moles of NaOH by multiplying the volume in liters (0.015L) by the molarity of 0.1

Answer 44

So that gives me 0.015moles of NaOH added

Answer 45

with me so far?

Answer 46

So when we added 0.015moles of NaOH, we reached the equivalence point. And we know that at the equivalence point, the moles of weak acid equals the moles of strong base (b/c they neutralize each other completely to form just the conjugate base and water)

Answer 47

So this means there must have been 0.015 moles of weak acid to start with

Answer 48

agree?

Answer 49

Yep sorry, I was afk. That makes sense.

Answer 50

grr this is hard (not "basic" haha)

Answer 51

Yeah. I'm pretty much fed up with chem. Do you know where to go from there? Since I kind of already had that lol

Answer 52

sec still writing (i used to be sharp at this but haven't done gen chem in a while)

Answer 53

ok i just found something awesome

Answer 54

Alrightyyy

Answer 55

the initial concentration of acid must be 0.1M b/c all of it was neutralized by 15ml of 0.1M of NaOH b/c the equivalence point was reached afterwards

Answer 56

Thus, the concentration of conj base at the equivalence point is 0.1M. Also, since the pH of the equivalence point is 8.6, we can half that to get the half equivalence point of 4.3. And since pH = pKa at the half equivalence point, the pKa of the acid is 4.3

Answer 57

half of the equivalence point isn't equal to half of the pH. The half equivalence point is determined by the negative log of Ka

Answer 58

@Abhisar , Sorry to ask, but could I use your help again?(:

Answer 59

so are u sure we can't just divide the pH at the equivalence point by 2 to get the half equivalence point?

Answer 60

Nooo, that's not how it works lol.

Answer 61

Ya...I'm starting to think there's isn't enough information in this question

Answer 62

I set up a RICE table for the reaction HA + H2O <=> H3O+ + A-

Answer 63

You mean ICE table, I'm pretty sure RICE is raise, ice, compress and elevate lmfao

Answer 64

ICE is Initial, Change, Equilibrium

Answer 65

Then I set up an equilibrium expression of Ka = [H3O+][A-]/[HA], and after substitution I got Ka = X^2/(a-x), where a is the initial concentration of HA

Answer 66

I know, the R just stands for Reaction

Answer 67

And x stands for the [H+]

Answer 68

We know the [H+] b/c we're given an initial pH of 3.2 so the [H+] concentration is 6.31E-4 M and that's also x in our equilibrium expression

Answer 69

Alright...

Answer 70

So after substituting for x, we get Ka = (6.31E-4M)^2 / (a - 6.31E-4M). I'm going to assume a is much larger than 6.31E-4M so we can further simplify this expression to Ka = (6.31E-4M)^2 / a

Answer 71

we need to find either Ka or a (which is the initial concentration of the acid)

Answer 72

so that's one equation....

Answer 73

Another equation is the M1V1 = M2V2 equation

Answer 74

I think you're off the mark. We've never used that equation with titration.

Answer 75

M1 is the initial concentration of the acid (or "a"), which we don't know, and V1 is the initial volume of acid, which we don't know either. M2 is the initial concentration of NaOH which is 0.1M and V2 is the initial volume of NaOH which is 0.015L.

Answer 76

What we do know is that 0.015L of 0.1M of NaOH, or in other words, 0.0015 moles of NaOH is equal to M1V1

Answer 77

so we have M1V1 = 0.0015 moles of acid

Answer 78

and that's all I can really get equation wise from this problem

Answer 79

This isn't your typical titration problem. I feel like there's a lack of information so you can't solve it but if I'm wrong and it's solvable, this is a super hard problem and you have to think outside the box

Answer 80

Okay, well I converted that concentration to get the pH and that's not one of the answers available.

Answer 81

What are your answer choices?

Answer 82

And what do you mean you converted that "concentration"? There's no concentration to convert here lol

Answer 83

We're looking for the pH not the molar concentration. My options are 8.6, 7.2, 3.4, 5.1

Answer 84

well i got 3.27, which is closest to 3.4

Answer 85

The questions asks for an "approximate" pH so what I did was "approximated" the pKa by halving the equivalence point to get an approximate pKa of 4.3

Answer 86

(I know that's not the proper way to do it)

Answer 87

So since pKa = -logKa, the Ka is 5.01E-5

Answer 88

I plugged that into the equation I stated before:

Ka = x^2 / (a - x), and after substitution I get

5.01E-5 = (6.31E-4)^2 / (a - 6.31E-4), and I get a = 8.58E-3, so that's our approximate initial concentration of weak acid

Answer 89

I plugged that into M of acid x V acid = M of base x V base, after substituting I get
8.58E-3 x V acid = 0.1M NaOH x 0.015L NaOH and my V acid is 0.175L

Answer 90

Alright. I don't quite understand the whole problem, but it's 4:21 in the morning so I'm gonna go with that lol

Answer 91

So my moles of acid are M acid x V acid, which is 8.58E-3 x 0.175, which is 0.0015. The last sentence says 5ml of 0.1M NaOH are added, which is the same thing as 0.0005 moles of NaOH. I subtract (neutralize) the moles of NaOH added from the amount of moles of acid present so I get 0.0015 moles acid - 0.0005 moles NaOH = 0.001 moles of acid remaning. Then I divide the remaining 0.001 moles of acid by the volume 0.175L and you got a Molarity of 5.71E-3M of acid. If you set up a RICE table with the acid's new concentration of 5.71E-3M, you get Ka = [H3O+][A-]/[HA]

Answer 92

After substituting, you get 5.01E-5 = x^2 / 5.71E-3, and x = [H3O+] = 5.35E-4. So if you negative log 5.35E-4, you get 3.27, which is the approximate pH!

Answer 93

Lol that's all I got for you, sorry

Answer 94

Thankyou for all your hard work, I really appreciate it. Hopefully it is the right answer(:

Answer 95

My pleasure. I hope it is too! P.S. let me know what the right answer was after your teacher tells you and how to solve it too please ^_^

Answer 96

I really want to know how you're properly supposed to solve this lol

Answer 97

Anyways, good night (or technically "morning" lol)