Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to
(a) 35.0 mL of 0.250 M NaOH(aq).
(b) 15.0 mL of 0.350 M NaOH(aq).
3 answers
Some one please help
I found the answer to a however i keep doing something wrong for b
My guess is that you didn't take into account the total volume.
millimols HCl=mL x M = 25 x 0.250 = 6.25
mmols NaOH = 15 x 0.35 = 5.25
NaOH + HCl ==> H2O + NaCl
mmols HCl = 6.25
mmols NaoH = 5.25
difference is excess HCl by 1.00 mmol.
So you have 1.00 mmols HCl in a total volume of 40 mL.
M HCl = mmols/mL = ? and convert H^+ to pH.
Post your work if you still don't get the right answer.
millimols HCl=mL x M = 25 x 0.250 = 6.25
mmols NaOH = 15 x 0.35 = 5.25
NaOH + HCl ==> H2O + NaCl
mmols HCl = 6.25
mmols NaoH = 5.25
difference is excess HCl by 1.00 mmol.
So you have 1.00 mmols HCl in a total volume of 40 mL.
M HCl = mmols/mL = ? and convert H^+ to pH.
Post your work if you still don't get the right answer.
2 answers