Question

4. What is the molar concentration and grams/Liter of a NaOH solution if 86 ml are titrated to an

endpoint by 375 ml of a solution of HCl that is .0175 M?
g/L: ________
Molarity: ____

Answer 1

0.76 M

30 g/L

Step-by-step explanation:

Step 1: Given data

• Molarity of the acid (Ma): 0.175 M

• Volume of the acid (Va): 375 mL

• Molarity of the base (Mb): ?

• Volume of the base (Vb): 86 mL

Step 2: Calculate the molarity of the base

We will use the following expression.

`Ma * Va = Mb * Vb\\Mb = (Ma * Va)/(Vb) = (0.175M * 375mL)/(86mL) = 0.76 M`

Step 3: Calculate the concentration of the base in g/L

The molar mass of NaOH is 40.00 g/mol.

`(0.76mol)/(L) * (40.00g)/(mol) = 30 g/L`