Help please!!! pH nightmare!!?

Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 2.78-M NaOH is added to 630. mL of each of the following solutions.



a) water



pH before mixing = 7

pH after mixing= 12.64

pH change = 5.64



b) 0.148 M NH41+



pH before mixing = ?

pH after mixing= ?

pH change = ?



c) 0.148 M NH3



pH before mixing = ?

pH after mixing= ?

pH change = ?



d) a buffer solution that is 0.148 M in each NH41+ and NH3



pH before mixing = ?

pH after mixing= ?

pH change = ?



I found a but am having a lot of trouble figuring out how to do b c and d. I have no clue.Help please!!! pH nightmare!!?
just keep in mind that when doing dilutions that C1*V1 = C2*V2 so if you are adding 10 ml of 2.78 mol/L NaOH to 630 mL of water and you want to find the new concentration so you can determine the pH, you can just plug in.



(2.78M)(10mL)=(C2)(630mL)

C2 = .004M



with that new conc you can find the pH

you must find the pOH since you know the conc of how much acid you are adding, not base

pOH = -log[OH-]

where [OH-] is the concentration you just solved for

kind in mind that pH + pOH = 14



note, this only works for assumed near 100% ionizationHelp please!!! pH nightmare!!?
uh, yea im not gunna do your homework for you.

grey hair

ireland