A 0.100 mole quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA

is the value of for the acid HA.

Explanation:

The pH of the solution = 3.75

The hydrogen ion concentration :

Concentration of weak acid = 0.100 mole/L

Initually

0.100 M    0   0

At equilibrium

(0.100-x)M     x    x

The expression of dissociation constant can be given as:

Here the value of x =

is the value of for the acid HA.

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the answer would be a