Chemistry

100 ml of 0.01N HCl is added to 0.9L of distilled water. What would the final pH of the solution be?

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asked by Kristin
  1. 0.01N HCl x (100 mL/1000 mL) = ?
    Then pH = -log(HCl).
    Note: This assumes the volumes are additive. Strictly speaking they are not; however, I think the intent of the question is to assume they are additive.

    posted by DrBob222

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