Calculate the ph of a 3.78×10-3 m solution of naf, given that the ka of hf = 6.80 x 10-4 at 25°c.

The answer is pH = 7.37.

Solution:
First, we set up an ICE table for the reaction:                                            
                          F- + H2O → HF + OH-
     Initial             0.00378         0        0
     Change        -x                   +x      +x
     Equilibrium   0.00378-x      x        x

We can calculate Kb from the given Ka, since we know that Kw = Ka*Kb = 1.0 x 10^-14 at  25°C:
     Kb = Kw/Ka = 1.0x10^-14 / 6.80x10^-4 = 1.471 x 10^-11
     Kb = 1.471 x 10^-11 = [OH-][HF] / [F-] = (x)(x) / (0.00378-x)

Approximating that x is negligible compared to 0.00378 simplifies the equation to
     1.471x10^-11 = (x)(x)  / 0.00378          
     1.471 x 10-11 = x2 / 0.00378 

Then we solve for x that is also equal to [OH-]:
     x2 = (1.471 x 10^-11)(0.00378)   
     x = sqrt[(1.471 x 10^-11)(0.00378)] = 2.358x10^-7 = [OH-]
in which 0.0000002358 is indeed negligible compared to 0.00378.

We can now calculate for pOH:         
     pOH = -log [OH-] = -log (2.358x10^-7) = 6.63

Therefore, the pH is
     pH = 14 - pOH = 14 - 6.63 = 7.37

Density because it’s holyyy