# You make two liters of a 3.0 pH solution of H2SO4:

You make 2 liters of a 3.0 pH solution of H2SO4:

1. how much water will i need to add to make the pH4.0 / 5.0?

2. How many more times more dilute is a 1.0 pH solution than the original?

3. What is the concentration of [h+] in the 1.0 pH solution and in the 3.0 ph solution?

4. If you have one mole of H2SO4, How many hydrogen ions are there? sulfate ions?

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mle219 wrote:If they told you it's a strong acid, you really would not need to know which acid it was to answer this part.

A strong acid is completely dissociated into ions in the pH range of 1.0 to 5.0.

Also, in that pH range, the amount of H

^{+}ions contributed by water will be insignificant compared to the amount coming from the acid.As a consequence, you can assume that all the H

^{+}in solution comes from H_{2}SO_{4}, and that all the H in H_{2}SO_{4}is in solution as H^{+}ions, before, and after diluting.You only need to know the concentrations of H

^{+}ions.Since pH = -log[H

^{+}], [H^{+}] = 10^{-pH}So, for pH = 3.0, H

^{+}] = 10^{-3.0}M = 0.001MFor pH = 4.0 and 5.0, you can calculate [H

^{+}] similarly.Once you know the concentrations before and after, you can easily calculate how much to dilute.

For example, if you needed to dilute to 0.0001M, since the pH = 3.0 solution is 0.001M, you would be diluting 10 times. The final volume would be 20 liters. You would have to add 18 more liters.

KMST

Thu, 2011-06-02 21:07

mle219 wrote:What do you mean by the original? Is it the pH=3.0 solution?

And what is a 1.0 pH solution? Did you mean a solution 1.0 higher in pH?

A pH=1.0 solution would be much more concentrated than a pH=3.0 solution.

Within limits, each increase of 1.0 units in pH means a solution ten times more dilute. The "limits" are in that at less than pH=1.0, even a strong acid may not be quite 100% dissociated, and that at pH>5 the H

^{+}and OH^{-}ions contributed by water would matter, and you could never get to pH>7 just by diluting an acid.mle219 wrote:In a solution with pH=1.0, [H

^{+}], the concentration of H^{+}, would be 0.1M,pH = -log[H

^{+}], and[H

^{+}] = 10^{-pH}KMST

Thu, 2011-06-02 21:27