Question

If the NaOH we use is 99.2% pure, how much NaOH should you weigh out to make 1 liter of 0.100 M NaOH? _______ g 99.2% NaOH

Calculate the molarity for a 0.1 M NaOH solution that was standardized by titrating 0.8023 g of KHP (potassium hydrogenphthalate) with 39.24 mL of NaOH solution.

Answer 1

To make a 0.100 M NaOH solution from 99.2% pure NaOH, 4.03 grams of the substance is required. After standardizing with KHP, the molarity of the NaOH solution was calculated to be approximately 0.1001 M.

Step-by-step explanation:

To prepare a 1 liter 0.100 M NaOH solution from 99.2% pure NaOH, we first determine the mass of pure NaOH needed. The number of moles required is:

0.100 mol/L × 1 L = 0.100 mol NaOH

From the molar mass of NaOH (40.00 g/mol), the mass of pure NaOH is:

0.100 mol × 40.00 g/mol = 4.00 g NaOH

Accounting for the purity, the mass of 99.2% pure NaOH needed is:

4.00 g NaOH / 0.992 = 4.03 g 99.2% NaOH

Standardizing the NaOH Solution with KHP;

To calculate the molarity of the standardized NaOH solution:

1. Find the moles of KHP reacting with NaOH. Molar mass of KHP (KHC8H4O4) is approximately 204.22 g/mol.
2. Moles of KHP = 0.8023 g / 204.22 g/mol = 0.00393 mol.
3. Since the stoichiometry of KHP reacting with NaOH is 1:1, moles of NaOH = moles of KHP.
4. Calculate the molarity of the NaOH solution: Moles NaOH / Volume of NaOH solution in liters.
5. 0.00393 mol / 0.03924 L = 0.1001 M NaOH.

The standardized NaOH solution is approximately 0.1001 M.