Question

Based

Answer 1

1) [H+] = [HCOO-]= x
[HCOOH] = 0.1-x

Ka = 1.78 x 10^-4 = [H+][HCOO-] / [HCOOH] = (x)(x) / 0.1-x

x = [H+]= 0.00422 M
pH = - log 0.0042 =2.37

2)    [H+] = [F-]= x
[HF] = 0.1-x

Ka =3.53 X10^-4 = [H+][HF-] / [HF] = (x)(x) / 0.1-x

x = [H+]= 5.94X10^-3 M
pH = - log 0.0059 = 2.23

3)    [H+] = [HCOO-]= x
      [H2CO3] = 0.1-x

Ka =4.45X10^-7 = [H+][HCOOO-] / [H2CO3] = (x)(x) / 0.1-x

x = [H+]= 2.11X10^-4 M
pH = - log 2.11x10^-4 = 3.68

4)    [NH4+] = [OH-]= x
       [NH3] = 0.1-x

       Kb = 1.8 x 10^-5 = [NH4+][OH-] / [NH3] = (x)(x) / 0.1-x

       x = [OH-]= 4.2X10^-4 M
pOH = - log 4.2X10^-4 =3.37

pH = 14-3.37= 10.63

5)   since 1 mole of H2SO4 releases 2 moles of H+
0.01 Molar H2SO4 releases 0.02 Molar H+

pH = - log of [H+]
pH = - log of [0.02]
pH = 1.7

pH 0.01M HCOOH= 2.37,   pH 0.10M HF=2.23 ,   pH 0.01M H2CO3 =3.68, pH 0.01M H2SO4= 1.7 ,pH 0.1M NH4+= 10.63

the order is NH4>H2CO3>HCOOH>HF>H2SO4