acids and bases question
nitric acid, HNO3, and nitrous acid, HNO2 , are 2 bronsted-lowry acids containing nitrogen. a student measures the pH of 0.045 mol dm-3 solutions of HNO3 and HNO2. (pKa = 3.35) and found the acids had different pHs.
a) why are the pH different?
b) calculate the pH value of 0.045 mol dm-3 HNO3
c) calculate the pH value of 0.045 moldm-3 HNO2
I can't answer these questions? with parts b and c, I don't know how i'm meant to change the equation
a) why are the pH different?
b) calculate the pH value of 0.045 mol dm-3 HNO3
c) calculate the pH value of 0.045 moldm-3 HNO2
I can't answer these questions? with parts b and c, I don't know how i'm meant to change the equation
1) HNO3 is a strong acid so it fully dissociates H+ ions solution while HNO2 will partially dissociate H+ ions in solution.
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
Original post by Giovannii
1) HNO3 is a strong acid so it fully dissociates H+ ions solution while HNO2 will partially dissociate H+ ions in solution.
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
ah I see! thanks. I wasn't aware HNO2 was a weak acid :/
Original post by Giovannii
1) HNO3 is a strong acid so it fully dissociates H+ ions solution while HNO2 will partially dissociate H+ ions in solution.
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
I just had a look at the OCR A spec to try and work out if there were any weak acids i need to learn off by heart. I couldn't see any. I only know things like CH3COOH is weak.
Original post by A*my
I just had a look at the OCR A spec to try and work out if there were any weak acids i need to learn off by heart. I couldn't see any. I only know things like CH3COOH is weak.
In the OCR A spec they mostly want you to know that:
HCl, HNO3, H2SO4, HBr, HI, and HClO4 are all strong acids.
Most other acids will be weak acids they will give it away that it is a weak acid by giving you a Ka or pKa.
Original post by Giovannii
1) HNO3 is a strong acid so it fully dissociates H+ ions solution while HNO2 will partially dissociate H+ ions in solution.
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
2) In a strong acid [H+] = [HA] so you will only need to -log (H+) to get the pH
3) For a weak acid first, you work out Ka by 10^pKa. Then using that value, H+^2 will be ka x [HA] but pH is -log(H+). So you would need to square root H+ then -log it
Hope this helps
how do you get the [H+]
For b and c
B.) conc of HNO3 is given as 0.450moldm-3 so just -log[0.450] which gives a PH of 0.35
C.) conc of HNO2 is given as 0.450 so
[HA] = 0.450
Pka of HNO2 is given as 3.35 so to find Ka it’s 10-3.35 which is 4.46x 10*-4
Ka = [H+] [A-] / [HA] And [H+] = [A-] so
Ka = [H+]*2 / [HA]
To find [H+] = square root (Ka x [HA] )
= square root( 4.46x10*-4 x 0.450)
= 0.014
Ph = -log [0.014]
Ph = 1.85
!!!
B.) conc of HNO3 is given as 0.450moldm-3 so just -log[0.450] which gives a PH of 0.35
C.) conc of HNO2 is given as 0.450 so
[HA] = 0.450
Pka of HNO2 is given as 3.35 so to find Ka it’s 10-3.35 which is 4.46x 10*-4
Ka = [H+] [A-] / [HA] And [H+] = [A-] so
Ka = [H+]*2 / [HA]
To find [H+] = square root (Ka x [HA] )
= square root( 4.46x10*-4 x 0.450)
= 0.014
Ph = -log [0.014]
Ph = 1.85
!!!
(edited 5 months ago)
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