"Calculate the pH of NaCl".
I know that NaCl is neutral...but I don't know how to show that it is neutral. The ka value for the Cl ion is "very large", so what do I do?
3 answers
Do for what? What's the question?
Sorry.
"Calculate the pH for 0.10 mol/L aqueous solution of NaCl".
"Calculate the pH for 0.10 mol/L aqueous solution of NaCl".
I thought I remembered this from two/three nights ago. I have tried hints and more hints and it hasn't worked so I'll work the problem.
NaCl is the salt of a strong base (NaOH) and a strong acid (HCl); therefore, neither the Na^+ nor the Cl^- will hydrolyze in water solution (the only salts that will hydrolyze are those that will produce a weak acid or weak base or both). So all we have is a little salt in water and the ion product of water holds.
H2O ==> H^+ + OH^-
We call (H^+) = x
If (H^+) is x, then (OH^-) = x
We know that (H^+)(OH^-) = Kw = 1 x 10^-14.
(x)(x) = 1 x 10^-14
x2 = 1 x 10-14
x = sqrt(1 x 10-14
x = 1 x 10-7 Molar.
pH = -log(H^+)=-log(1 x 10-7) = 7.0
NaCl is the salt of a strong base (NaOH) and a strong acid (HCl); therefore, neither the Na^+ nor the Cl^- will hydrolyze in water solution (the only salts that will hydrolyze are those that will produce a weak acid or weak base or both). So all we have is a little salt in water and the ion product of water holds.
H2O ==> H^+ + OH^-
We call (H^+) = x
If (H^+) is x, then (OH^-) = x
We know that (H^+)(OH^-) = Kw = 1 x 10^-14.
(x)(x) = 1 x 10^-14
x2 = 1 x 10-14
x = sqrt(1 x 10-14
x = 1 x 10-7 Molar.
pH = -log(H^+)=-log(1 x 10-7) = 7.0