Question

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

Answer 1

Answer:

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:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base

• For a:

We are given:

$n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=9.50\times 10^{-2}M=0.00950M\\V_2=50.0mL$

Putting values in above equation, we get:

$1\times 0.105\times V_1=1\times 0.00950\times 50\\\\V_1=\frac{1\times 0.00950\times 50}{1\times 0.105}=4.52mL$

Hence, the volume of HCl needed is 4.52 mL

• For b:

We are given:

$n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.117M\\V_2=23.0mL$

Putting values in above equation, we get:

$1\times 0.105\times V_1=1\times 0.117\times 23\\\\V_1=\frac{1\times 0.117\times 23}{1\times 0.105}=25.63mL$

Hence, the volume of HCl needed is 25.63 mL

• For c:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

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:

$\text{Molarity of NaOH solution}=\frac{1.40}{40\times 1}\\\\\text{Molarity of NaOH}=0.035M$

We are given:

$n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.035M\\V_2=130mL$

Putting values in above equation, we get:

$1\times 0.105\times V_1=1\times 0.035\times 130\\\\V_1=\frac{1\times 0.035\times 130}{1\times 0.105}=43.33mL$

Hence, the volume of HCl needed is 43.33 mL