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Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80. You have in front of you 100 mL of 7.00×10−2 M HCl, 100 mL of 5.00×10−2 M NaOH, and plenty of distilled water. You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 84.0 mL of HCl and 88.0 mL of NaOH left in their original containers. Part A Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

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Answer:

15.1 mL of the HCl solution should be added

Explanation:

First off, let's use the definition of pH to find the desired number of H⁺ moles:

  • pH = -log[H⁺]
  • 2.80 = -log[H⁺]
  • [tex]10^{-2.8}[/tex] = [H⁺]
  • [H⁺] = 1.58 x 10⁻³ M
  • Then we use the definition of Molarity [C=n/V]:
  • [H⁺] = 1.58 x 10⁻³ M = molesH⁺ / 1.00 L
  • molesH⁺ =  1.58 x 10⁻³

So in the end there needs to be 1.58 x10⁻³ moles of H⁺ to achieve the desired pH.

Now let's calculate the added amounts of HCl and NaOH

  • mL HCl = 100 - 84 = 16 mL = 0.016 L
  • mL NaOH = 100 - 88 = 12 mL = 0.012 L
  • mol HCl = 0.016 L * 7.00x10⁻² M = 1.12x10⁻³ mol HCl = mol H⁺
  • mol HCl = 0.012 L * 5.00x10⁻² M = 6.0x10⁻⁴ mol NaOH = mol OH⁻

OH⁻ reacts with H⁺, producing water. So the moles of H⁺ remaining in the solution are:

  • 1.12x10⁻³ - 6.0x10⁻⁴ = 5.2x10⁻⁴ mol H⁺

This means that the moles of H⁺ that we need to add are:

  • 1.58x10⁻³ - 5.2x10⁻⁴ = 1.06x10⁻³ mol H⁺

Finally we calculate the required volume of HCl:

  • C = n/V
  • 7.00 x10⁻² M = 1.06x10⁻³ mol H⁺ / V
  • V = 0.0151 L = 15.1 mL

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