Question

How many liters of 1.0 M HCl do you need to neutralize 2.0 L of 3.0 M NaOH? How many liters of 1.0 M HCl do you need to neutralize 1.5 L of 3.0 M Ca(OH)2? How many liters of 1.0 N H2SO4 do you need to neutralize 1.5 liters of 3.0 N Sr(OH)2?

Answer 1

Answer:

For 1: The volume of HCl required is 6 L.

For 2: The volume of HCl required is 9 L.

For 3: The volume of sulfuric acid required is 4.5 L.

Explanation:

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

$n_1M_1V_1=n_2M_2V_2$        ......(1)

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 1:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L$

Putting values in equation 1, we get:

$1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L$

Hence, the volume of HCl required is 6 L.

• For 2:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L$

Putting values in equation 1, we get:

$1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L$

Hence, the volume of HCl required is 9 L.

• For 3:

To calculate the volume of acid, we use the equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of acid

$N_2\text{ and }V_2$ are the normality and volume of base

We are given:

$N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L$

Putting values in above equation, we get:

$1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L$

Hence, the volume of sulfuric acid required is 4.5 L.