NaOH + HCl ==> NaCl + H2O
moles NaOH initially = in the problem.
moles HCl initially = M x L = ??
Subtract the two (the smaller from the larger) and the difference is the amount of the larger one in excess. Then
H^+ or OH^- = (moles larger/total L) = ??
Convert that answer to pH.
Post your work if you get stuck.
What is the pH of the solution formed when 0.040mol NaOH(s) is added to 2.00L of 0.020M HCl?
4 answers
So initial mol NaoH= 0.040mol
and initial HCl= (.02M)(2.00L)= 0.04mol
I don't understand what to do now.
and initial HCl= (.02M)(2.00L)= 0.04mol
I don't understand what to do now.
Usually these don't come out the same and I didn't do the calculation in my head. Probably I should have.
Since moles HCl = moles NaOH, neither is in excess and you have prepared some NaCl in water.
So you have, except for the NaCl, which won't change the pH, just water. Water, as you know, has a pH = 7. You can calculate that if you wish.
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1 x 10^-14
(H^+)^2 = 1 x 1o^-14
(H^+) = sqrt(1E-14) = 1 x 10^-7 and pH = 7.
Since moles HCl = moles NaOH, neither is in excess and you have prepared some NaCl in water.
So you have, except for the NaCl, which won't change the pH, just water. Water, as you know, has a pH = 7. You can calculate that if you wish.
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1 x 10^-14
(H^+)^2 = 1 x 1o^-14
(H^+) = sqrt(1E-14) = 1 x 10^-7 and pH = 7.
OH! I understand now. Thanks! I didn't understand why the answer was pH 7 before.
1 answer