CALCULATING pH?

Question

calculate the pH of 1.0 x10^-10 M NaOH

Answer 1

NaOH is a strong base and dissociate entirely according:

NaOH-----> Na^+  + OH^-1

so the concentration of OH^-1 ions is equal to that one of NaOH

pOH= - log c(OH^-1)mol/L/mol/L

pOH= - log 1.0 x10^-10 mol/L/mol/L

pOH=10

pKw=pH+pOH       pKw=14

14=pH+10

pH=4

So, the people who corrected me are entirely right, pH cannot be 4, but almost neutral, it is so obviously, but I get that pH=4, didnt I :)))

Answer 2

Correct, the pH CANNOT be 4, adding a base to water does not make it more acidic, that is contradictory.  Since Na)H is a strong base, the pH would have to be above 7.0, but not very much.  The pH can be calculated using the systematic treatment of equilibria:

1.  Write the charge balance

2.  Write the mass balance

3.  Write any relevant equilibria

4.  Solve for the unknown

For NaOH in water, we have two reactions:

1.  NaOH ---> Na+ + OH-     this goes to completion because NaOH is a strong base.

2.  H2O ---> H+ + OH-     this does NOT go to completion, so we need the equilibria expression for water:

3.  Kw = [H+][OH-] = 1.0e-14

Mass Balance:  [Na+] = 1.0e-10 M because that is what you added as NaOH, and it completely dissocates, so all Na+ is equal to the NaOH put in.

Charge Balance:  The concentration of all positive ions must equal the concentration of all negative ions.  We have Na+, H+ and OH- in the solution so:

     [Na+] + [H+] = [OH-]

Equilibria: [H+][OH-] = 1.0e-14

We have 3 equations and 3 unknowns, so we can solve the problem.  Since we want to determine [H+], make is x for simplicity.

1.  [Na+] + x = [OH-],     so     (1.0e-10 + x) = [OH-]

2.  (x) (1.0e-10 + x) = 1.0e-14

3.  x^2 + (1.0e-10)(x) - 1.0e-14 = 0

We have a quadratic equation which can be solved using

     x = (-b +/- SQRT(b^2 - 4ac)) / (2a)

With some math, we end up with :

x= (-1e-10 +/- sqrt(1.0e-20 + 4e-14) ) / 2

x = (-1e-10 +/- 2e-7) / 2

x = 9.995e-8  and x = -1.0005e-7

Since we can not have a negative concentration,

x = [H+] = 9.995e-8

and pH = -log[H+]

so pH = 7.0002

Answer 3

If in the solution there is nothing else but H2O and NaOH, the NaOH concentration is so low that  in all practical purposes the pH is almost neutral, that is just a little above 7.  But it cannot be 4 which would need the presence of an acid