Original post by Jshek
I cant do 3a , so 1.30×10^-3 =[ H^+] and [0.1]
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
ka = [H+][A-]/[HA]
[H+] = [A-]
ka = [H+]2/[HA]
0.1 x 1.3 x 10-3 = [H+]2
[H+] = root[1.3 x 10-4]
[H+] = 0.0114 mol dm-3
pH = 1.94
Original post by charco
ka = [H+][A-]/[HA]
[H+] = [A-]
ka = [H+]2/[HA]
0.1 x 1.3 x 10-3 = [H+]2
[H+] = root[1.3 x 10-4]
[H+] = 0.0114 mol dm-3
pH = 1.94
[H+] = [A-]
ka = [H+]2/[HA]
0.1 x 1.3 x 10-3 = [H+]2
[H+] = root[1.3 x 10-4]
[H+] = 0.0114 mol dm-3
pH = 1.94
Cheers
Original post by Jshek
I cant do 3a , so 1.30×10^-3 =[ H^+] and [0.1]
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
What is the full question? Is it a strong or weak acid?
Original post by Jshek
Cheers
Why did you ask the same question again?
Original post by Jshek
I cant do 3a , so 1.30×10^-3 =[ H^+] and [0.1]
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
I then solve for H+
Then - log base 10 H+ which doesnt give me 1.94
It’s a weak acid so you can’t just convert H+ to pH.
[H+] = (square root) Ka/[HA]
Then you can convert to pH
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