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What is the pH of a solution resulting from 5.00 mL of 0.011 M HCl being added to 50.00 mL of pure water?

User Wesleyy
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Answer:

Assuming that the addition of 5.00 mL of 0.011 M HCl to 50.00 mL of pure water does not significantly affect the volume of the solution:

Calculate the number of moles of HCl added:

moles HCl = concentration x volume = 0.011 mol/L x 0.00500 L = 5.50 x 10^-5 mol

Calculate the total volume of the solution:

total volume = 50.00 mL + 5.00 mL = 55.00 mL = 0.055 L

Calculate the concentration of H+ ions in the solution:

[H+] = moles HCl / total volume = 5.50 x 10^-5 mol / 0.055 L = 1.00 x 10^-3 M

Calculate the pH of the solution:

pH = -log[H+] = -log(1.00 x 10^-3) = 3.00

Therefore, the pH of the solution is 3.00.

Step-by-step explanation:

User Ced
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