if 50 ml of a solution of ph 3 is mixed with 50 ml of a solution of 5, explain why the resulting ph is still about 3 and not 4?
Dear Abdulhafiz
the no of mole of the 50ml of ph3 solution= (1x10^-3)x (50/1000)
= 5x10^-5
the number of moles in the 50ml of ph5 solution= (1x10^-5) x (50/1000)
= 5x10^-7
Total no of moles in the solution
= 5x10^-5 + 5x10^-7
= 5.05x10^-5
The molarity of the new soltion
= (5.05x10^-5) / (100/1000)
= 5.05x10^-4
The ph value
(ph = -log H+)
ph = -(log5.05x10^-4)
ph= 3.30
You must find the mol of the new solution so that you are able to find the ph value of the new solution, rather then just finding the average ph value of the new solution(since it will not be accurate)
Hope it helps!