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AnnZ [28]
3 years ago
14

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 o

f 3.
Chemistry
1 answer:
Ksivusya [100]3 years ago
4 0

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

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