Question

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to neutralize this basic solution. What is the molarity of the HCl solution?
1. 0.015 M2. 0.043 M3. 0.0037 M4. 0.040 M5. 0.060 M

Answer 1

Answer: The correct answer is Option 5.

Explanation:

• To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

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

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$  

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

• To calculate the molarity 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 which is $HCl$

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

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.