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1. Came across this past paper Q and I really don't know how to go about doing it!
Mark scheme is v. concise, but the answer is 19.4g

How would you get this ans though?

I've attached the Q

would appreciate it soooo much!
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2. (Original post by Jasmine_777)
Came across this past paper Q and I really don't know how to go about doing it!
Mark scheme is v. concise, but the answer is 19.4g

How would you get this ans though?

I've attached the Q

would appreciate it soooo much!
In the previous must they must have told you which one is the acid and which one is the salt!
well if they didn't then we must work it out..this is not OCR is it!

EDIT:I got as far as finding the ratio of salt/acid..then I need a volume which is not given..
3. (Original post by Jasmine_777)
Came across this past paper Q and I really don't know how to go about doing it!
Mark scheme is v. concise, but the answer is 19.4g
The weak acid equilibrium:

Ka = [H+][A-]/[HA]

where [HA] is the acid, [A-] is its conjugate base.

You are told that pH = 7, therefore [H+] = 1 x 10-7

Ka = 6.2 x 10-8

Both compounds are in the same solution so volumes cancel out. Hence:

6.2 x 10-8/1 x 10-7 = 0.1/[HA]

[HA] = 0.1613

Relative mass = 120

Therefore mass needed in 1dm3 = 0.1613 * 120 = 19.35g

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