Question

A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The end point was obtained at about 24.8 mL. The concentration of the monoprotic acid is about ........ mol/L.
A) 25.0

B) 0.0600

C) 0.240

D) 0.120

E) None of the above

Answer 1

Answer:

The concentration of the acid is about 0.114 M (option E)

Explanation:

Step 1: Data given

Volume of the monoprotic acid = 25.0 mL = 0.025 L

Molarity of the monoprotic acid = ?

Molarity of the NaOH solution = 0.115 M

Volume NaOH = 24.8 mL = 0.0248 L

Step 2: Calculate the concentration

a*Cb * Vb = b * Ca * Va

⇒ a = the coeficient of NaOH = 1

⇒ Cb = the molarity of the acid = TO BE DETERMINED

⇒ Vb = the volume of the acid = 0.025 L

⇒ b = the coefficient of the acid = monoprotic = 1

⇒ Ca = the moalrity of NaOH = 0.115 M

⇒ Va = the volume of NaOH = 0.0248 L

1 * Cb * 0.025 = 1 * 0.115 * 0.0248

0.025 Cb = 0.002852

Cb = 0.11408 M

The concentration of the acid is about 0.114 M