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.
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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