Answer:
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
[tex]6.58x10^{-4}\frac{g}{L} /62\frac{g}{mol} = 1.06x10^{-5}\frac{mol}{L}[/tex]
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
[tex]K_{a} = 4.3x10^{-7} = \frac{x^{2}}{1.06x10^{-5}-x }[/tex]
Clearing the x, and solving the quadratic equation
[tex]4.3x10^{-7}.[1.06x10^{-5}-x] = x^{2}\\4.6x10^{-12} - 4.3x10^{-7}x - x^{2} = 0\\x1 = 1.9x10^{-6} \\x2=-2.4x10^{-6}[/tex]
Only the positive value is correct, so [H+ ]=[HCO3 − ] = 1.9x10-6
Since, pH = - Log [H+ ]
[tex]pH = - Log [H^{+} ] = -Log (1.9x10^{-6}) = 5.7[/tex]
