128 ml of 1.50 x 10-2 m naoh is used to titrate 225 ml of a monoprotic acid with unknown concentration to reach the equivalence point. what is the concentration of the original acid?

1.50 x 10-2 m

1.17 x 10-1 m

1.92 x 10-3 m

8.53 x 10-3 m

M = 8.53x10⁻³ M

Explanation:

In this case, we are having an acid base titration, and the reaction taking place is the following:

NaOH + HA <> H₂O + NaA

This equation shows that the mole ratio between the acid and the base is 1:1 therefore we can say:

nA = nB (1)

If the expression to calculate moles is:

n = M*V

we can replace it in (1), and then, solve for concentration of the acid. This way:

MA*VA = MB*VB

Replacing the data:

225MA = 128 * 1.5x10⁻²

MA = 128 * 1.5x10⁻² / 225

MA = 8.53x10⁻³ M

So the correct option is the last one.