How many milliliters of 0.105 M HCl are needed to titrate each of the following solutions to the equivalence point?

a. 50.0 mL of 9.5010?2 M NaOH

b. 23.0 mL of 0.117 M NH3

c. 130 mL of a solution that contains 1.40 g of NaOH per liter

For a: The volume of HCl needed is 4.52 mL

For b: The volume of HCl needed is 25.63 mL

For c: The volume of HCl needed is 43.33 mL

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid

are the n-factor, molarity and volume of base

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl needed is 4.52 mL

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl needed is 25.63 mL

To calculate the molarity of solution, we use the equation:

Given mass of NaOH = 1.40 g

Molar mass of NaOH = 40 g/mol

Volume of solution = 1 L

Putting values in above equation, we get:

We are given:

Putting values in above equation, we get:

Hence, the volume of HCl needed is 43.33 mL

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