If the volume of the solution is changed then how will its pH change?

initially there is 100 ml of 0.01M NaOH solution which is diluted to 1L then find the initial and the final pH of the solution......if anyone can solve this question????If the volume of the solution is changed then how will its pH change?
Find the initial pH of the NaOH solution (strong base).

pOH= -log [OH-]= -log[0.01]= 2

pH= 14-pOH= 14-2= 12



Solve for moles NaOH:

(100 mL= 0.1 L)

0.01M NaOH= x moles NaOH/ 0.1 L

x moles NaOH= 0.001 mol NaOH



Now solve for new concentration with 1 L (NOTE: Moles NaOH does not change):

0.001 mole NaOH/ 1 L= 0.001 M



Solve for new pH:

pOH= -log [OH-]= -log[0.001]= 3

pH= 14-pOH= 14-3= 11



Easy way: pH works on logorithmic scale. When you dilute something by 10 times (100 mL to 1000 mL), you're moving down by one pH.

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