[Sette]

What will be the pH of a solution made up by mixing 1 litre of an acid solution of pH 4.2 with 2 litres of another acid solution with a pH of 6.0?

Ive tried a few ways, firstly by averaging the two, by the amount actually in the solution, but im sure this is not right.

The other was taking the stronger pH mixture of which there is more and using that.

Any suggestions of help be appreciated thx.

[Borek]

Assuming these are solutions of strong acids calculate moles amount of H+ in both, add, and divide by the total volume.

[Sette]

WOuld it be same formula for calculation if a base and acid mixed?

[Borek]

No - when you mix acid and base they react to produce water and salt (it is called neutralization) and pH depends on what was in excess - acid or base.

[Sette]

Still having trouble working out my initial question, after following your advice i got something which didnt look right.

6.5 x 10^-8

[Borek]

Show details of your calculations.

[Dude]

Borek is correct.

pH = -log(H+)
-4.2 = log H+ =  =0.00006309 mol H/L
-6 = log H+  = 0.000001 mol H/L

2 L (conc of 6 pH solution 0.000001 mol H / L) + 1 L (0.00006309 mol H/L) / 3L
= 0.0000216... mol H / L
- log (0.0000216 mol H/L) = 4.7 pH

Buffering effects will make prediction of the H+ concentration more difficult.  The ideal case is that the H are from strong acid solutions.

[Borek]

Dude: I asked for Sette calculations, not yours