Question

A certain weak base has a Kb of 7.70 × 10-7. What concentration of this base will produce a pH of 10.03?

Answer 1

Let us first write the equation,

A +H2O ------> HA +OH-

Let initIal concentration be A be "A "

Let "x" be the amount of concentration reacted

A    +     H2O ------> HA        + OH-

A                                   -              -

A-x                               x                 x

now pH desired is 10.03

pH + pOH =14

pOH = 14 - pH = 14-10.03 = 3.97

Converting pOH to [OH-]

pOH = - log( [OH-])

3.97 = - log( [OH-])

[OH-] = 10 ^-3.97 =1.071519 *10^-4 M

x =[OH-] = [HA] = 1.071519 *10^-4 M

Let us write Kb expression,      

   

Kb = 7.70*10^-7

Substituting the values of Kb and concentration in terms of x

we know x= 1.071519 *10^-4 M, substituting it in the above equation

          7.70 *10^-7 (A -1.071519*10^-4) = (1.071519*10^-4) ²

                  (A -1.071519*10^-4)   = (1.071519*10^-4) ² / 7.70 *10^-7 = 0.014911

                  A - 1.071519*10^-4 = 0.014911

                A = 0.014911 + 1.071519*10^-4 = 0.015018 M

A concentration of 0.015018 M will produce a pH of pH 10.03