A 0.2 M carboxylic acid (RCOOH) has a Ka = 1.66x10-6. What is the pH of this solution? Enter to 2 decimal places.

3.24

Explanation:

The dissociation equation for the carboxylic acid can be represented as follows:

RCOOH —-> RCOO- + H+

We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.

RCOOH RCOO- H+

Initial 0.2 0.0. 0.0

Change -x +x. +x

Equilibrium 0.2-x. x. x

We can now find the value of x as follows:

Ka = [RCOO-][H+]/[RCOOH]

(1.66* 10^-6) = (x * x)/(0.2-x)

(1.66 * 10^-6) (0.2-x) = x^2

x^2 = (3.32* 10^-7) - (1.66*10^-6)x

x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0

Solving the quadratic equation to get x:

x = 0.0005753650094369094 or - 0.0005753650094369094

As concentration cannot be negative, we discard the negative answer

Hence [H+] = 0.0005753650094369094

By definition, pH = -log[H+]

pH = -log(0.0005753650094369094)

pH = 3.24

i do not get it   at all