Question

Chromic acid, H₂CrO₄ is a diprotic acid, with pKa values of 0.86 and 6.51. Calculate the total alkalinity for a water sample containing 10^-3 M Na₂CO₃ and 10^-4 M H₂CrO₄

Answer 1

Concentration of H2CrO4 = 10^-4M

pKa1 = 0.86

Ka1 = 0.138

Ka1 = [ HCrO4-] [ H+]/ [H2CrO4]

X^2 = 0.138 × (0.0001)

=1.38 × 10^-5

X = 3.71×10^-3

[H+] = 3.71× 10^-3M

[HCrO4-] = 3.71 × 10^-3M

pK2 = 6.51

Ka2 = 2.75 × 10^-7

Ka2 = [ CrO42- ] [ H+]/[HCrO4-]

X^2 = 2.75×10^-7 × 3.71 × 10^-3

= 3.19 × 10^-5

[H+] = 3.19× 10^-5M

[CrO4^2- ] = 3.19 × 10^-5MM

Na2CO3 -------> 2Na+ + CO3^2-

CO3^2- + H2O <------> HCO3- + OH-

Kb = Kw/Kacid

= 1× 10^-14/4.8×10^-11

=2.08 ×10^-4

Kb = [ HCO3-][OH-]/[CO3^2-]

X^2 =2.08 × 10^-4× 10^-3

X = 4.5×10^-4

[ OH-] = 4.5 × 10^-4

[ HCO3-] = 4.5 × 10 ^-4

Total alkalinity =([OH-] + [HCO3-] +[ 2 × [CO3^2- ]] +[HCrO4-] + [2 × CrO4^2-]) - [H+]