Question

Calculate the pH of a 0.45M solution of Benzonic Acid. (Use the quadratic equation to solve).

Answer 1

Answer – Given, [C₆H₅COOH] = 0.45 M , Ka for the C₆H₅COOH = 6.4*10^-5

C₆H₅COOH + H₂O ------> H₃O⁺ + C₆H₅COO^-

I   0.45                               0           0

C    -x                                 +x          +x

E   0.45-x                            +x          +x

Ka = [H₃O⁺] [C₆H₅COO^-] / [C₆H₅COOH]

6.4*10^-5 = x*x /(0.45-x)

6.4*10^-5 *(0.45-x) = x²

Now we need to set up quadratic equation

2.88*10^-5 - 6.4*10^-5 x = x²

x^{2 +} 6.4*10^-5 x - 2.88*10^-5 = 0

a =1 , b = 6.4*10^-5, c = -2.88*10^-5

Using the quadratic equation

x = -b +/- √b²-4a*c / 2a

Plugging the value in this formula

x = 0.00533 M

so, x = [H₃O⁺] = 0.00533 M

so, pH = -log [H₃O⁺]

           = -log 0.00533 M

           = 2.27

So, the pH of a 0.45M solution of Benzonic Acid is 2.27