Assume you prepared a solution by adding 0.05mL (1 drop) of 0.5M HCl to 5x10^7mL

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Explanation / Answer

I would do this but check my thinking. (HCl) added = 0.5M x (0.05/5E7)= 5E-10M ..............H2O ==> H^+ + OH^- initial...............0......0 change.................x......x add 5E-10M...........5E-10....x final..............x+5E-10.....x Kw = (H^+)(OH^-) = 1E-14 (x+5E-10)(x)=1E-14 Multiply out to obtain a quadratic that looks like this. x^2 + 5E-10x - 1E-14 = 0 Solve the quadratic and I get something like 9.975E-8 = x. (I know that's more significant figures than we can legally use; however, if we round everything down to one place (from the 0.5M and the 0.05 mL) it would be 7 anyway and there would be no need to do any of this calculation. Then x+5E-10 = ?? and pH from there. Close to 7 but less than 7 as it should be. It looks logical anyway.

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