A10.0 ml solution of 0.300 m nh3 is titrated with a 0.100 m hcl solution. calculate the ph after the addition of 10.0 ml hcl.

pH = 9.55

[NH3] = 2.0 mmol/20.0 mL = 0.100 M

[NH4+] = 1.00 mmol/20.0 mL = 0.0500 M

Ka = (1.0 × 10^-14)/(1.8 × 10^-5)

     = 5.56 × 10^-10

pKa = log Ka = 9.25

pH = pKa + log ([base]/[acid])

    = 9.25 + log (0.100/0.0500)

    = 9.55

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