Jshek
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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
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Jshek
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Jshek
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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
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charco
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(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
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
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Jshek
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(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
Cheers
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Sdecicco
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(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
What is the full question? Is it a strong or weak acid?
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charco
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(Original post by Jshek)
Cheers
Why did you ask the same question again?
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Sdecicco
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(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
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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