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

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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%