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

Answer:

Volume of water added = 2.0 L

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,

$M_1V_1=M_2V_2$

$0.002 \times 2.0 = 0.001 \times V_2$

$V_2 = \frac{0.002 \times 2.0}{0.001} = 4.00\ L$

Volume added = Final volume - Initial volume

                        = 4.0 - 2.0 = 2.0 L