Chemistry Questions (1 Viewer)

[albertcamus]

Re: Does anyone have detailed notes on these two experiments.

a) Seems like a graph/data based question.
b) Seems like a graph/data based question also - sounds like the Haber Process though, which is in Module 3 (however it could be given as a question regarding LCP - I can't say given what you've said.) But not 100% sure - although I think this is from the option topic Industrial Chemistry.
c) Not 100% sure - but I think this is from the option topic Industrial Chemistry.
d) Not 100% sure - but I think this is from the option topic Industrial Chemistry.
e) Definitely from the option topic Industrial Chemistry.

Actually I think all of this might be from Industrial Chemistry...strange to see it in a half yearly paper...

 

[girlworld_club]

Curious... what would happen to the PH of the salt solution if an aqueous salt solution of NaOH is distilled with 500g of water. Would it increase, decrease or remain the same?

 

[HeroicPandas]

Curious... what would happen to the PH of the salt solution if an aqueous salt solution of NaOH is distilled with 500g of water. Would it increase, decrease or remain the same?

Well NaOH is a base, dilution will result in a decrease in pH

 

[girlworld_club]

A 250ml sample of 0.10 mol/L HCL is addded to 250ml of 0.025mol/L NaOH. Calculate the final pH of the mixture.

 

[girlworld_club]

also what is the reaction between sodium hydroxide + hydronium ion ?

 

[bleakarcher]

A 250ml sample of 0.10 mol/L HCL is addded to 250ml of 0.025mol/L NaOH. Calculate the final pH of the mixture.

HCl(aq)+NaOH(aq)->NaCl(aq)+H2O(l)
n(HCl)=0.1*0.25=0.025 mol
n(NaOH)=0.025*0.25=0.00625 mol
Hence, the limiting reagent will be the aqueous NaOH solution since we have less moles of it and because for every mole of HCl that is neutralised one mole of NaOH is neutralised.
After the complete neutralisation of NaOH, we have 0.025-0.00625=0.01875 mol HCl in an aqueous solution of volume 250+250 mL= 0.5 L. So the concentration of HCl will be 0.01875/0.5=0.0375 mol/L. Since HCl is a strong acid, it ionises completely i.e. HCl(aq) -> H+(aq) + Cl-(aq), and so the concentration of H+ ions will be equal to concentration of HCl i.e. [H+]=0.0375 meaning pH(of the final solution)=-log([H+])=-log(0.0375)=3.28.

Note that NaCl is a neutral salt so I neglected it in the final pH calculation.

 

[girlworld_club]

thanks everyone for your responses. Another question:
If you have a beaker of 500mL of a solution with a pH of 4 and THEN add 500 mL of WATER to it, what is the pH of the final 1.0 L that is left.