Question

Find the ph for 0.350M of RbF

Answer 1

Solution :-

RbF is the conjugate base of the HF

Therefore using the Kb of the F^- we can calculate the pH of the solution

     F^-    +    H2O   --------- >    HF + OH^-

    0.350 M                             0                0

    -x                                      +x              +x

    0.350-x                               x                x

Kb= [HF][OH-]/[F^-]

1.47*10^-11 = [x][x]/[0.350-x]

since the Kb is very small therefore we can neglect the x from the denominator then we get

1.47*10^-11 = [x][x]/[0.350]

1.47*10^-11 * 0.350 = x^2

5.15*10^-12=x^2

taking square root of both sides we get

2.27*10^-6 = x =[OH-]

now lets calculate the pOH

pOH= -log [OH-]

pOH = -log [2.27*10^-6]

pOH = 5.64

pH + pOH = 14

therefore pH = 14 - pOH

                  = 14 - 5.64

                  = 8.36

Therefore the pH of the solution is 8.36