Question

Determine the amount of water that must be added to a 2-litre solution of sulphuric acid to dilute it from a pH of 2.7 to a pH of 3.

Answer 1

Volume of water added = 2.0 L

Step-by-step explanation:

Initial pH of the solution = 2.7

[H^+] concentration in 2 L solution of sulfuric acid,

`pH = -log[H^+]`

`[H^+] = 10^(-pH)\ M`

`[H^+] = 10^(-2.7) = 0.001995\ M = 0.002\ M`

Final pH of the solution = 3

Final [H^+] concentration of sulfuric acid,

`[H^+] = 10^(-pH)\ M`

`[H^+] = 10^(-3) = 0.001\ M`

Now,

`M_1V_1=M_2V_2`

`0.002 * 2.0 = 0.001 * V_2`

`V_2 = (0.002 * 2.0)/(0.001) = 4.00\ L`

Volume added = Final volume - Initial volume

= 4.0 - 2.0 = 2.0 L