# A-level chemistry help urgenttt!!!

#1
So I have the following question and iv been stuck on it for a while and iv got a whole lot of them but I don't know how to do it

Calculate the PH of the solutions formed in the following way.
Addition of 250 cm3 of water to 50cm3 of 0.2 moldm-3 HN03

0
2 years ago
#2
wouldnt it remain the same
0
#3
Why would it remain the same?
Sorry I'm just trying to understand
0
2 years ago
#4
because H2O splits into H+ and OH- which both cancel eachother out
0
2 years ago
#5
are u given the ka/pka in the question?
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2 years ago
#6
calculate the change in the concentration of HNO3. now that you have a new concentration and HNO3 is a strong acid, the pH of the solution is the negative log (to the base ten) of the concentration of H+ ions, which, in this case, is the same as the concentration of the acid
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#7
So it would just be -log(0.2)? I'm still confused though ahaha wouldn't adding water dilute it hence changing the ph?
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#8
(Original post by ComicMage031)
calculate the change in the concentration of HNO3. now that you have a new concentration and HNO3 is a strong acid, the pH of the solution is the negative log (to the base ten) of the concentration of H+ ions, which, in this case, is the same as the concentration of the acid
How would I work out the change in concentration? Sorry do you think you could talk me through it please?
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2 years ago
#9
yeah so we first calculate the total number of moles in the 50cm3 solution which is 0.05*0.2 and equals 0.01 moles of HNO3. Now we calculate the new concentration with the new volume, (50 + 250 = 300). to do that we just divide the number of moles, 0.01, by the volume in DM3, 0.3. the new concentration comes to be (1/30) or (0.033). now we calculate the concentration of H+ ions in our 0.0333M solution. since HNO3 is a strong acid, that is it disassociates completely when dissolved in water, the concentration of H+ ions in the solution is the same as the concentration of HNO3; 0.0333 moldm-3. now we take the negative log of the concentration of H+ ions. -log[H+] = -log[0.03333] = 1.477 ~ 1.48
1
#10
Awesome thanks
1
#11
Well we're here, I have another question
To calculate the ph of 10 gdm-3 Hcl
So I have done: 10/36.5 = 0.27
-log(0.27) = 0.56
Is this correct?
0
2 years ago
#12
(Original post by SAB10)
Well we're here, I have another question
To calculate the ph of 10 gdm-3 Hcl
So I have done: 10/36.5 = 0.27
-log(0.27) = 0.56
Is this correct?
I haven't got my calculator out to check your working, but the method is right.
0
2 years ago
#13
Well, this just put me off A-level chem
0
1 year ago
#14
Heyyy
I had the same question and that is what I did!
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