Question

•You have 560. mL of aluminum hydroxide, with a molarity of 0.0300 M. When mixed with 871 mL of hydrochloric acid, what is the molarity (M) of the acidic solution?

•How many liters of 0.488 M H3PO4 are produced with 35.1 mL of 0.320 M of Ca(OH)2 ?

Answer 1

For 1: The molarity of acidic solution is 0.579 M

For 2: The volume of acidic solution is 0.0153 L

Step-by-step explanation:

To calculate the concentration 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 1:

We are given:

`n_1=1\\M_1=?M\\V_1=871mL\\n_2=3\\M_2=0.030M\\V_2=560mL`

Putting values in above equation, we get:

`1* M_1* 871=3* 0.0300* 560\\\\M_1=(3* 0.0300* 560)/(1* 871)=0.579M`

Hence, the molarity of acidic solution is 0.579 M

• For 2:

We are given:

`n_1=3\\M_1=0.488M\\V_1=?L\\n_2=2\\M_2=0.320M\\V_2=35.1mL=0.0351L`

Putting values in above equation, we get:

`3* 0.488* V_1=2* 0.320* 0.0351\\\\V_1=(2* 0.320* 0.0351)/(3* 0.488)=0.0153L`

Hence, the volume of acidic solution is 0.0153 L