Question

A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in the lake water ?

Answer 1

The molarity (M) of the acid in the lake water is $0.00001M$ .

Explanation:

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$  are molarity of acid in the lake water and base solution respectively.$V_1,V_2$  are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

$M_1=0.00001M$

Therefore, the molarity (M) of the acid in the lake water is $0.00001M$ .