1) The initial equation can be rearranged to give: $$[H^+] =
10^{-pH}$$
The pH of water is 7, and so: $$[H^+] = 10^{-7}mol/l.$$
Thus in 1l of water there are 10$^{-7}$ moles of hydrogen
ions.
2) If the pH is 2, then $[H^+] = 10^{-2} mol/l$
If this is diluted to a volume of 2 litres from the initial 1
litre, then $[H^+] = 0.5 \times 10^{-2}mol/l$.
$$\therefore pH = -log_{10}(0.5 \times 10^{-2}) = 2.30\ (2\
d.p)$$
3) If pH = 1.3, then $[H^+] = 10^{-1.3} mol/l$
Therefore, in 100ml (= 0.1l), there are $0.1 \times 10 ^{-1.3}$
moles.
We want to dilute this such that the pH = 2, which is the same as
$[H^+] = 10^{-2} mol/l$.
Therefore, since we have already 100ml of solution, this requires
the addition of 401ml.
4) For the first acid, pH = 3, therefore $[H^+] = 10^{-3}
mol/l$.
Since there is 400ml of this acid, then the number of $H^+$ is $0.4
\times 10^{-3}$ moles
For the second acid, pH = 4, therefore $[H^+] = 10^{-4}
mol/l$.
Since there is 300ml of this acid, the number of $H^+$ is $0.3
\times 10^{-4}$ moles
If these are mixed, there are a total of $4.3 \times 10^{-4}$ moles
of $H^+$ in a volume of 0.7l. Therefore, $$[H^+] = 6.14 \times
10^{-4} mol/l$$