01-21-2007, 10:05 PM
Evrerything is matter of this formula: pH= -log[H+]
Pure Water which is neutral has pH= -log[concentration in H+]= - log[0,0000001]= -[-7]=7
Solution of 2mMol has a concentration in H+ equal to 0.002 so pH= -log[0.002]
But went it got dilutated with same amount fo pure water its concentration fell down to half of intial which is 0.002/2= 0.001
So the pH of the new solution is -log[0.001] = 3
So the pure water of pH of 7 switch to pH= 3 when they added to it the solution.
NB log stands for decimal logarithm
Thanks
Pure Water which is neutral has pH= -log[concentration in H+]= - log[0,0000001]= -[-7]=7
Solution of 2mMol has a concentration in H+ equal to 0.002 so pH= -log[0.002]
But went it got dilutated with same amount fo pure water its concentration fell down to half of intial which is 0.002/2= 0.001
So the pH of the new solution is -log[0.001] = 3
So the pure water of pH of 7 switch to pH= 3 when they added to it the solution.
NB log stands for decimal logarithm
Thanks