Question

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

Answer 1

Answer:

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.

Answer 2

M1V1 = M2V2
128(.015) = 225M2
M2 = 8.53x10^-3 or D