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# unit 4 ph calculations watch

1. http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
q5c

mol of acid=0.00375
mol of base=0.006

and that's as far as i got. Can't remember how to go any further.
2. How many mol of acid or alkali remain after they've reacted.

Don't forget the molar ratio of the reaction.

Don't forget when calculating the concentration of the one in excess that you've got a new volume.

The rest should follow.
3. (Original post by Pigster)
How many mol of acid or alkali remain after they've reacted.

Don't forget the molar ratio of the reaction.

Don't forget when calculating the concentration of the one in excess that you've got a new volume.

The rest should follow.
Excess base0.006-0.00375=0.00225

new conc 1000x0.00225/30+25 = 0.041

H+=1x10^-14 / 0.041 = 2.4444444444444x10^-13

-logH+ = 12.6
4. (Original post by will'o'wisp)
Excess base0.006-0.00375=0.00225

new conc 1000x0.00225/30+25 = 0.041

H+=1x10^-14 / 0.041 = 2.4444444444444x10^-13

-logH+ = 12.6
2 mol of KOH react with 1 mol of H2SO4
5. (Original post by Pigster)
2 mol of KOH react with 1 mol of H2SO4
so where do i add the double ratio in?
6. (Original post by will'o'wisp)
so where do i add the double ratio in?
How about this: rather than worrying about the amount of acid and of the alkali, do the calculation in terms of the amount of H+ and OH-.
7. (Original post by Pigster)
How about this: rather than worrying about the amount of acid and of the alkali, do the calculation in terms of the amount of H+ and OH-.
I don't know how to do that?
8. 1 mol of KOH makes 1 mol of OH-

1 mol of H2SO4 makes 2 mol of H+
9. (Original post by Pigster)
1 mol of KOH makes 1 mol of OH-

1 mol of H2SO4 makes 2 mol of H+
so was it supposed to be 0.0075 mol fo acid then?
10. (Original post by will'o'wisp)
so was it supposed to be 0.0075 mol of acid then?
Effectively, yes.
11. (Original post by Pigster)
Effectively, yes.
o i see thanks then, thta it for today ... gotta get on with philosophy...

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