Question

An old bottle labeled Standardized 5.0 M NaOH was found at the back of a shelf in the stockroom. To determine whether the concentration was still 5.0 M, 5.0 mL of the solution was diluted to 100 mL and titrated to the equivalence point with 11.9 mL of 2.0 M HCl(aq). What is the molarity of the sodium hydroxide solution in the bottle

Answer 1

Answer: The molarity of the sodium hydroxide solution in the bottle is 0.238 M

Step-by-step explanation:

To calculate the volume 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 which is NaOH.

We are given:

`n_1=1\\M_1=2.0M\\V_1=11.9mL\\n_2=1\\M_2=?\\V_2=100mL`

Putting values in above equation, we get:

`1* 2.0* 11.9=1* M_2* 100\\\\M_2=0.238`

Thus the molarity of the sodium hydroxide solution in the bottle is 0.238 M