Calculate the concentration of all species in a 0.160 m solution of h2co3.

             H₂CO₃ ⇔       HCO₃⁻   +     H⁺

I            0.160               0                 0

C            -x                  +x               +x

E          0.160-x          +x                +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x)   (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

             HCO₃⁻    ⇔            CO₃⁻²    +   H⁺

I          2.62 x 10⁻⁴               0                2.62 x 10⁻⁴

C          -x                            +x              +x

E       2.62 x 10⁻⁴ - x           +x              2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

2b2t

explanation:

2b2t

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.