If the kb of a weak base is 3.3 ã 10-6, what is the ph of a 0.17 m solution of this base?

The relation between base equilibrium constant and degree of dissociation is

C x^2 = Kb

x is degree of dissociation

C x^2 = 3.3 * 10^-6

(0.17) x^2 = 3.3 * 10^-6

x = 4.24 * 10^-3

Concentration of hydroxyl ion is

[OH-] = c x = 0.17 * 4.24 * 10^-3 = 0.72 * 10^-3 M

PoH = - log [OH-] = 2.85

PH = 14 - pOH = 11.15

Hence, pH of solution is 11.15

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml