Question

how does the consentration of hydrogen ions changes when the solution of an acid is diluted with water?

Answer 1

Let’s take an example base which can be said to be fully dissociated in water, e.g. NaOHNaOH. You take 40 g40 g of the pure compound and dissolve it in0.5 l0.5 l of water. The standard way to calculate the concentration of sodium ion is:

c=nV=mMV=40 g40 g⋅mol−1×0.5 l=2 moll(1)(1)c=nV=mMV=40 g40 g⋅mol−1×0.5 l=2 moll

This builds on the fact that the amount of sodium ions is identical to the amount of NaOHNaOH added. The same thing can be done for hydroxide ions; within experimental error the concentration of hydroxide ions is also 2 mol⋅l−12 mol⋅l−1 as calculated in equation (1)(1).

Now say you add another 0.5 l0.5 l to the solution you generated above. Obviously, you did not add any additional sodium ions, so their amount will remain the same. The new concentration can be calculated as shown in (2)(2):

c′=nV′=mMV′=40 g40 g⋅mol−1×1.0 l=1 moll(2)(2)c′=nV′=mMV′=40 g40 g⋅mol−1×1.0 l=1 moll

Thus, in the resulting solution, you are left with half the original concentration of sodium ions; similarly for hydroxide ions at first approximation. The concentrationdecreased by adding water