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+ ]

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

answerca air mass;