Chemistry 3/4 2013 Thread

[LOLs99]

I have a question and this looks complicated.

Question: One litre of an aqueous solution of potassium hydroxide(KOH) has a PH of 12.0 at 25C.

The amount of pure HCL gas, in mol ,that must be added to the solution to lower the PH from PH 12.0 to 2.0 would be

A 10

B 2.0

C 0.02

D 0.01

[thushan]

I have a question and this looks complicated.

Question: One litre of an aqueous solution of potassium hydroxide(KOH) has a PH of 12.0 at 25C.

The amount of pure HCL gas, in mol ,that must be added to the solution to lower the PH from PH 12.0 to 2.0 would be

A 10

B 2.0

C 0.02

D 0.01

[OH-]initially = 0.01 M
n(OH-)initially = 0.01 mol (1 L)

n(H+)final = 0.01 mol.

Therefore need 0.02 mol of H+ - 0.01 mol to neutralise OH-, and 0.01 mol as excess.

[LOLs99]

Wow, that was a quick response
I dont get how do u get those numbers.

[thushan]

Wow, that was a quick response
I dont get how do u get those numbers.

No worries!

pH = -log 10 [H+]

So pH = 12 means [H+] = 10^-12. From self-ionisation of water, we know [H+][OH-]=10^-14. So [OH-]=10^-2 M.
Or pH + pOH = 14, so pOH = 2, so [OH-]=10^-2 M.

We know you have 1 L of solution, so the amount of OH- is 0.01 mol.

Now, pH final is 2, meaning [H+] after reaction is 10^-2 M. Since we still have 1 L of solution, amount of H+ after reaction is 0.01 mol.

So how much HCl is required to decrease pH from 12 to 2?

So, we need 0.01 mol of HCl to react with the 0.01 mol of OH-, and another 0.01 mol of HCl to provide the 0.01 mol of H+.

So 0.02 mol.

[LOLs99]

No worries!

pH = -log 10 [H+]

So pH = 12 means [H+] = 10^-12. From self-ionisation of water, we know [H+][OH-]=10^-14. So [OH-]=10^-2 M.
Or pH + pOH = 14, so pOH = 2, so [OH-]=10^-2 M.

We know you have 1 L of solution, so the amount of OH- is 0.01 mol.

Now, pH final is 2, meaning [H+] after reaction is 10^-2 M. Since we still have 1 L of solution, amount of H+ after reaction is 0.01 mol.

So how much HCl is required to decrease pH from 12 to 2?

So, we need 0.01 mol of HCl to react with the 0.01 mol of OH-, and another 0.01 mol of HCl to provide the 0.01 mol of H+.

So 0.02 mol.

Ahh, great explanations . I guess I get how to do this sort of questions now.

 More ques(Help me to check if my answers are correct) :
1. Mass of hydrated copper(II) sulfate sample -8.81g
    Mass of residue after heating- 5.41g

Use these results to determine the mass of water in the sample of hydrated copper(II) sulfate and from this information calculate the empirical formula of the sample.

MY ANSWER IS 2CuSO4 . 11H2O

2. Calculate the percentage by mass of sodium chloride in the sausage roll sample if an 8.45g sample of the sausage roll was analysed using the previous method and a 0.636g percipitate of silver chloride was obtained.

MY ANSWER IS 3.07%

[Reckoner]

^Yep 

[LOLs99]

Cool thanks. I feel good after getting questions correct haha

[bucklr]

Hi there,
currently we are doing our gravimetric analysis SAC on determining sulfate as sulfate in fertiliser.
It involves:
dissolving the fertiliser (filtering out insoluble materials), adding HCl (known concentration), then BaCl2(Aq) to precipitate the SO4 ions, then we precipitate the Cl- ions with AgNO3(Aq).
I was wondering why do you add HCl to the SO4 solution? I just cannot see its relevance. I would have understood if you didn't add HCl, as it would have clearly been for a checking mechanism. Adding the HCl just seems to complicate things.

Many thanks.

[Stick]

This is what I found from a quick Google search:

"To keep the pH of the solution with the sample slightly acidic to eliminate possible interference from anions of weak acids. The acidic wash is to maintain the electrolyte concentration that promotes coagulation."

So I guess we were asked to add hydrochloric acid to try and make sure the proper precipitates are formed and that we don't precipitate something that we shouldn't. I'd be very interested to know the chemical reasoning and mechanics of how this works, however, just out of pure curiosity.

[thushan]

This is what I found from a quick Google search:

"To keep the pH of the solution with the sample slightly acidic to eliminate possible interference from anions of weak acids. The acidic wash is to maintain the electrolyte concentration that promotes coagulation."

So I guess I'm getting that we were asked to add hydrochloric acid to try and make sure the proper precipitates are formed and that we don't precipitate something that we shouldn't. I'd be very interested to know the chemical reasoning and mechanics of how this works, however, just out of pure curiosity.

CO2 in the atmosphere can dissolve into water and react with water to form H2CO3, which ionises to form HCO3- ions. Some may even ionise to form CO3 (2-) ions. These two ions can potentially precipitate out cations. Adding excess acid protonates these ions back to H2CO3, which is expelled as CO2.

[Stick]

Thanks. My lab partner and I couldn't figure it out.

[teletubbies_95]

Woah! Same situation as you. We did the same prac yesterday, and I was confused why we added HcL too! Thanks for the question and the amazing explanations!

[Limista]

Woah! Same situation as you. We did the same prac yesterday, and I was confused why we added HcL too! Thanks for the question and the amazing explanations!

Isn't Stick's question about why we add excess HCl, instead of which acid we add?
Wasn't TRex trying to answer why we have to add an excess of the precipitating reagent?

Sorry - I just want to make sure I have comprehend this properly.

[bucklr]

Isn't Stick's question about why we add excess HCl, instead of which acid we add?
Wasn't TRex trying to answer why we have to add an excess of the precipitating reagent?

Sorry - I just want to make sure I have comprehend this properly.

Stick was answering my question by saying an acidic solution helps the precipitation, TRex furthered his answer and explained why the solution needs to be acidic.

[LOLs99]

Ques:  A hydrocarbon on complete combustion gives a mass of carbon dioxide which is 20/9 times greater than the mass of water produced.  What is the empirical formula of the hydrocarbon?