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Solution A has a pH of 1, soultion B has a pH of 6. If I mix the same volume V of each solution into one, what will be the pH of the resulting solution?

I'm guessing here as no more information has been provided: It is a mixture of a strong acid with a weaker acid. So I'm assuming that the final concentration of protons will be related with the equilibrium displacement of the weaker acid's dissociation. The full dissociation of the stronger acid forces the equilibrium of the dissociation of the weaker to be displaced to the left, as in there will be no proton contribution from the weak acid. So the pH of the resulting solution will be computed from the concentration of protons from solution A being diluted in the new volume, 2V.

Am I in the right direction? I feel that calculating the raw amount of protons contributed by each solution and then computing the new concentration will be too obvious for a method.

Anything that would guide me into this pondering will be very much appreciated.

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    $\begingroup$ In the absence of any information as to what acids these are, the only real answer you can give is probably the "too obvious" one. $\endgroup$ – orthocresol Apr 19 '17 at 0:07

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