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Consider 54.0 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 4.200. What volume of water must be added to make the pH = 5.500?

I found HA for both and then I used C1V1 = C2V2 plugging in the concentrations of HA in C1 and C2 and 0.054 for V1, but I'm still not getting the right answer. Please help!

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