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

Answer:

For 1: The molarity of acidic solution is 0.579 M

For 2: The volume of acidic solution is 0.0153 L

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:

$1\times M_1\times 871=3\times 0.0300\times 560\\\\M_1=\frac{3\times 0.0300\times 560}{1\times 871}=0.579M$

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:

$3\times 0.488\times V_1=2\times 0.320\times 0.0351\\\\V_1=\frac{2\times 0.320\times 0.0351}{3\times 0.488}=0.0153L$

Hence, the volume of acidic solution is 0.0153 L