Question

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.

Answer 1

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

Step-by-step explanation:

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

`6.58x10^(-4)(g)/(L) /62(g)/(mol)  = 1.06x10^(-5)(mol)/(L)`

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

`K_(a) = 4.3x10^(-7)  = (x^(2))/(1.06x10^(-5)-x )`

Clearing the x, and solving the quadratic equation

`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)`

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

Since, pH = - Log [H+ ]

`pH = - Log [H^(+) ] = -Log (1.9x10^(-6)) = 5.7`