Calculate the ph of a water at 25°c that contains 0.6580 mg/l carbonic acid. assume that [h+ ]=[hco3 − ] at equilibrium and neglect the dissociation of water.

The pH of a water at 25°C that contains 0.6580 mg/L carbonic acid is 5.7.

Explanation:

First, the molar mass of carbonic acid is 62g/mol, thus

The solution has a concentration of 1.06x10-5 mol/L.

Second, the acid-base balance is araised, with the initial and the equilibrium concentrations

       H2CO3        +      H2O    ⇄      HCO3−     +      H3O+

i)      1.06x10-5                                        -                     -

eq)   1.06x10-5 - x                                 x                     x

Knowing the Ka of carbonic acid (4.3x10-7) the expression of the equilibrium constant is written

Clearing the x, and solving the quadratic equation

Only the positive value is correct, so [H+ ]=[HCO3 − ] = 1.9x10-6

Since, pH = - Log [H+ ]

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

2b2t

explanation:

2b2t