buffer ph question

so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below

buffer ph question.png156.5KB

Original post by dasda

so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below

Guess what?

Attachments do not show.

Reply 2

dasda

OP

15

Original post by charco

Guess what?

Attachments do not show.

Sheesh, I made this post at the same time as the other one
buffer ph question.png

Original post by dasda

Sheesh, I made this post at the same time as the other one
buffer ph question.png

"so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below"

But if the postassium hydroxide is reacting with the acid it makes more salt. How can it decrease?

Reply 4

dasda

OP

15

Original post by charco

"so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below"

But if the postassium hydroxide is reacting with the acid it makes more salt. How can it decrease?

ahh so the salt RcooK would increase. why would this cause a decrease in acid then.

Original post by dasda

ahh so the salt RcooK would increase. why would this cause a decrease in acid then.

"so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below"

You have already said why, yourself!

Reply 6

dasda

OP

15

Original post by charco

"so I thought that since pottasium hydroxide was neutralising the acid, equilibrium would shift to make more acid. so acid would increase by a certain amount and the salt would decrease by the same amount but I saw that this lie of thinking is wrong. I guessed the second part of the question and got it right but don't understand why it is right. attached below"

You have already said why, yourself!

Okay but I thought that if acid is neutralised, this would decrease the moles of acid. wouldn't equilibrium shift to make more acid.

Original post by dasda

Okay but I thought that if acid is neutralised, this would decrease the moles of acid. wouldn't equilibrium shift to make more acid.

Yes, but you need the acid dissociation equation to work that out:

ka = [H+][A-]/[HA]

and that's what bufffers are all about.

Reply 8

dasda

OP

15

Original post by charco

Yes, but you need the acid dissociation equation to work that out:

ka = [H+][A-]/[HA]

and that's what bufffers are all about.

I think I'M lost.

Original post by dasda

I think I'M lost.

OK

Information given:

0.0136 mol KX dissolved in 100 cm³ of 0.5M HX (weak acid)
3 x 10^-4 mol KOH added
ka = 1.41 x 10^-5

--------------------------------------------------------------------------------------------

3 x 10^-4 mol KOH reacts with 3 x 10^-4 mol of HX to make 3 x 10^-4 mol of KX

Final mol of HX = initial mol - reacted mol = 0.05 - 3 x 10^-4 = 0.0497 mol
Final mol of KX (X-) = initial mol + reacted mol = 0.0136 + 3 x 10^-4 = 0.0139 mol

ka = [H+][X-]/[HX]

although the equation shows concentrations all components are in the same volume, allowing us to use moles directly

[H+] = Ka[HX]/[X-] = 1.41 x 10^-5 x 0.0139/0.0497 = 3.94 x 10^-6

It's at the above step I'm not careful substituting the values and I put the HX moles and the KX moles in the wrong place! Doh!

So, I lose valuable marks in the exam.

[H+] = Ka[HX]/[X-] = 1.41 x 10^-5 x 0.0497/0.0139 = 5.04 x 10^-5

^{pH = 4.30}

(edited 3 years ago)

Reply 10

dasda

OP

15

Original post by charco

OK

Information given:

0.0136 mol KX dissolved in 100 cm³ of 0.5M HX (weak acid)
3 x 10^-4 mol KOH added
ka = 1.41 x 10^-5

--------------------------------------------------------------------------------------------

3 x 10^-4 mol KOH reacts with 3 x 10^-4 mol of HX to make 3 x 10^-4 mol of KX

Final mol of HX = initial mol - reacted mol = 0.05 - 3 x 10^-4 = 0.0497 mol
Final mol of KX (X-) = initial mol + reacted mol = 0.0136 + 3 x 10^-4 = 0.0139 mol

ka = [H+][X-]/[HX]

although the equation shows concentrations all components are in the same volume, allowing us to use moles directly

[H+] = Ka[HX]/[X-] = 1.41 x 10^-5 x 0.0139/0.0497 = 3.94 x 10^-6

pH = 5.40 (2 d.p.)

what about the second part of the question.

Original post by dasda

what about the second part of the question.

I've already explained it.

"although the equation shows concentrations all components are in the same volume, allowing us to use moles directly"

Reply 12

dasda

OP

15

Original post by charco

OK

Information given:

0.0136 mol KX dissolved in 100 cm³ of 0.5M HX (weak acid)
3 x 10^-4 mol KOH added
ka = 1.41 x 10^-5

--------------------------------------------------------------------------------------------

3 x 10^-4 mol KOH reacts with 3 x 10^-4 mol of HX to make 3 x 10^-4 mol of KX

Final mol of HX = initial mol - reacted mol = 0.05 - 3 x 10^-4 = 0.0497 mol
Final mol of KX (X-) = initial mol + reacted mol = 0.0136 + 3 x 10^-4 = 0.0139 mol

ka = [H+][X-]/[HX]

although the equation shows concentrations all components are in the same volume, allowing us to use moles directly

[H+] = Ka[HX]/[X-] = 1.41 x 10^-5 x 0.0139/0.0497 = 3.94 x 10^-6

pH = 5.40 (2 d.p.)