what volume of water is added to 400 ml of HCL solution so that the pH of the solution increases by 1 unit
harshit259:
plz help me.........
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Initial pH = x
After dilution, pH = x + 1
pH = - log[H+]
[H+] = 10^(-pH)
Initial concentration = 10^-x
Concentration after dilution = 10^-(x + 1)
From dilution law
M1V1 = M2V2
10^(-x) × 400 = 10^-(x + 1) × (400 + V)
10^(-x + x + 1) = (400) / (400 + V)
10 = (400 + V)/400
10 = 1 + V/400
9 = V/400
V = 3600 mL
∴ 3600 mL of water should be added to increase the pH by 1 unit.
If you’re unable to follow the calculations refer to the image attached.
After dilution, pH = x + 1
pH = - log[H+]
[H+] = 10^(-pH)
Initial concentration = 10^-x
Concentration after dilution = 10^-(x + 1)
From dilution law
M1V1 = M2V2
10^(-x) × 400 = 10^-(x + 1) × (400 + V)
10^(-x + x + 1) = (400) / (400 + V)
10 = (400 + V)/400
10 = 1 + V/400
9 = V/400
V = 3600 mL
∴ 3600 mL of water should be added to increase the pH by 1 unit.
If you’re unable to follow the calculations refer to the image attached.
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Answer:
Hey mate plzz refer to the attachment
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