The Kb of weak base B is 1.0 x 10-6. A solution contains 0.75 M B and 0.25 M HB+Cl- (the salt of B with HCl). What is the pH of the solution after 0.05 mol NaOH is added to 1.0 L of the above solution?

pH of the given solution is 8.7.

Explanation:

For the buffer mixture, initial pOH will be given as follows.

         pOH =

                 =

                 =

                 = 5.523

When 0.05 mol NaOH is added to this buffer solution then concentration of species present will be as follows.

   0.05 mol      0.25           0.75           0

      0              0.20      (0.75 + 0.25)   0.05

Hence, the volume of solution will be 1 liter.

      [BOH] = ,           [Salt] =

So,  pOH = [/tex]

                = 6 - 0.69

                = 5.30

Now, we will calculate the pH as follows.

            pH = 14 - 5.30

                  = 8.7

Thus, we can conclude that pH of the given solution is 8.7.

the answer would be a

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole