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 abov

Answer 1

The correct answer is c

Answer 2

Answer:

The concentration of the monoprotic acid is about 0.114 mol/L.

The correct answer is E none of the above

Explanation:

Step 1: Data given

Volume of a monoprotic acid = 25.0 mL = 0.025 L

Molarity of the NaOH solution = 0.115 M

The end point was obtained at about 24.8 mL.

Step 2: Calculate the concentration of the monoprotic acid

b*Ca*Va = a*Cb*Vb

⇒with B = the coefficient of NaOH = 1

⇒with Ca = the concentration of the monoprotic acid = ?

⇒with Va = the volume of the monoprotic acid = 0.025 L

⇒with a = the coefficient of the monoprotic acid = 1

⇒with Cb = the concentration of NaOH = 0.115M

⇒with Vb = the volume of NaOH= 0.0248 L

1*Ca*0.025 = 1*0.115*0.0248

Ca = (0.115*0.0248)/0.025

Ca = 0.114 M

The concentration of the monoprotic acid is about 0.114 mol/L.

The correct answer is E none of the above