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You are given 500 ml of 2N HCl and 500 ml of 5N HCl. The maximum volume (in ml) of the 3 M that you can make from these two solutions is ___________.
Correct Answer: 750
For HCl, 3 M is 3 N. Mixing $$V_1$$ of 2 N with $$V_2$$ of 5 N gives $$2V_1 + 5V_2 = 3(V_1 + V_2)$$, so $$V_1 = 2V_2$$. Only 500 mL of the 2 N acid is available, so $$V_1 = 500$$ and $$V_2 = 250$$, giving at most 750 mL.
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