# pH calculatiion

Submitted by kirachem on Tue, 02/02/2010 - 20:48

a) Determine the pH of 50mL of 0.20M Ca(OH)2?
b) Determine pH of 50.0mL of 0.20M NaOH + 50.0mL of 0.10M HCl?
c) Determine pH of 50.0mL of 0.20M NaOH + 50.0mL of 0.20M NaOH?
a) for a I went moles=(.05L)(0.20M)=0.01M
then I went pOH=-log(0.01M)=2
so pH=12? is this right can someone check for me
b) I went mole of NaOH=(0.05L)(0.20M)=0.01M
moles of HCl=(0.05L)(0.10M)=0.0050mol
but i dont know what to do from there
c) how do you do that if you have the same amount of NaOH added together?

pH is dependent on the concentration of H+ ions and not the number of ions. Concentration is expressed by molarity (M). When you multiply volume in L by molarity, you get the number of moles. At one point in your calculation, you have written a calculated number of moles as 0.01M but this should be 0.01 moles. (M is incorrect unit here).

a) each mole Ca(OH)2 contains 2 moles of OH-.

The volume of Ca(OH)2 is not important as the solution would have the same pH, whatever the volume.
So if [Ca(OH)2] = 0.20M, then [OH-] = 0.40M which is 0.40 mole/L
So plug this into pOH equation and then find pH

b) HCl  + NaOH  ---->  NaCl  + H2O
1 mol   1 mol
So 0.1 mole HCl will react with 0.1 mole NaOH. You have moreNaOH than  HCl, so some NaOH will be left unused.

Mole HCl = 0.005 moles, moles NaOH = 0.01. So 0.01 - 0.005 = 0.005 moles of NaOH will be left after neutralisation. This will be  in a volume of 100 mL so the concentration of NaOH and of OH- will be 0.05 moles/L. Plug this into pOH and then find pH.

c) You are mixiing two solutions together with the same concentration. So this will not change (only the volume). So concentration is still 0.20 moles/L. As the formula is NaOH, [OH-] is also 0.20 moles/L. Plug into pOH and then find pH