Chemsitry question help: Acid, Bases and pH
I have searched the internet for about an 1 hour and looked through my book so many times but I have some questions I have no idea how to answer. Could anyone just help me out?
How do I calculate the volume of a solution from using pH and 25.0cm3 sample of 0.0850 mol dm-3?
how do I calculate the pKa only using the temperature and Ka?
Any help would be nice, thanks x
How do I calculate the volume of a solution from using pH and 25.0cm3 sample of 0.0850 mol dm-3?
how do I calculate the pKa only using the temperature and Ka?
Any help would be nice, thanks x
Original post by ashpittawayx
I have searched the internet for about an 1 hour and looked through my book so many times but I have some questions I have no idea how to answer. Could anyone just help me out?
How do I calculate the volume of a solution from using pH and 25.0cm3 sample of 0.0850 mol dm-3?
How do I calculate the volume of a solution from using pH and 25.0cm3 sample of 0.0850 mol dm-3?
Your question makes no sense. You already have the volume of the solution!
how do I calculate the pKa only using the temperature and Ka?
Any help would be nice, thanks x
you don't need the temperature:
pKa = -log10Ka
Original post by charco
Your question makes no sense. You already have the volume of the solution!
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
you don't need the temperature:
pKa = -log10Ka
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
you don't need the temperature:
pKa = -log10Ka
Thanks I realised after I posted it x
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
how do i calculate the volume?
how do i calculate the volume?
Original post by ashpittawayx
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
how do i calculate the volume?
how do i calculate the volume?
you're really pretty
Original post by LisaNikita
you're really pretty
Thank you, so are you x
Original post by ashpittawayx
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
how do i calculate the volume?
how do i calculate the volume?
As you mentioned it earlier, do you have the Ka?
Original post by h3rmit
As you mentioned it earlier, do you have the Ka?
No, that was for a different question x
Okay. Firstly you know that there is initially 0.0850 moldm-3 of H+ ions. Using: (Volume x Concentration)/1000 you will get moles- Which in this case is 2.125 x 10-3 moles of H+ ions. This number of moles will stay the same after the distilled water is added. Using [H+] = 10-pH Find the concentration of H+ ions after the distilled water is added. Then rearrange the first equation: (moles x 1000)/Concentration and this will give you the final volume, using the concentration you just calculated and the moles you calculated before that. The volume will be in cm3.
Original post by ashpittawayx
this is my exact question: A 25.0cm3 sample of 0.0850 mol dm-3 hydrochloric acid was placed in a beaker. distilled water was added until the pH of the solution was 1.25. Calculate the volume of the solution formed. (3)
how do i calculate the volume?
how do i calculate the volume?
The concentration of the hydrogen ions is the same as the acid as it is a strong acid.
Work out the concentration of hydrogen ions in the target solution.
You know how many moles of acid you are starting with (and ending with) mol = MV
So now calculate the overall volume of solution with this number of moles which will give you the target concentration.
Ok so what you need to do:
Find moles of HCl initially in solution using cv/1000.
You know that [H+] and therefore [HCl] in the final solution is equivalent to the value for 10^-1.25.
Use the expression c=1000*moles/volume:
10^-1.25 = 1000 x moles/ 25 + x
Rearrange to find x.
Total volume is 25 + x.
If I'm wrong anybody please correct me
Find moles of HCl initially in solution using cv/1000.
You know that [H+] and therefore [HCl] in the final solution is equivalent to the value for 10^-1.25.
Use the expression c=1000*moles/volume:
10^-1.25 = 1000 x moles/ 25 + x
Rearrange to find x.
Total volume is 25 + x.
If I'm wrong anybody please correct me
Original post by hoafanuk
Ok so what you need to do:
Find moles of HCl initially in solution using cv/1000.
You know that [H+] and therefore [HCl] in the final solution is equivalent to the value for 10^-1.25.
Use the expression c=1000*moles/volume:
10^-1.25 = 1000 x moles/ 25 + x
Rearrange to find x.
Total volume is 25 + x.
If I'm wrong anybody please correct me
Find moles of HCl initially in solution using cv/1000.
You know that [H+] and therefore [HCl] in the final solution is equivalent to the value for 10^-1.25.
Use the expression c=1000*moles/volume:
10^-1.25 = 1000 x moles/ 25 + x
Rearrange to find x.
Total volume is 25 + x.
If I'm wrong anybody please correct me
Thank you, turns out I had already done half but didnt know where to go after, nice to know i would probs get some marks in the exam if it came up
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