Question

A student neutralizes 20.0 mL of a sodium hydroxide solution (NaOH) by adding 28.0 mL at 1.0 M hydrochloric acid (HCI). What is the molarity of the solution?

Answer 1

`=>(C_M)_(NaOH)=1.4M`

Explanation:

The net ionic equation is:

`H^+(aq)+OH^(-)(aq)->H_2O(l)`

Therefore, `n_(H^+)=n_(OH^(-)`

The relationship between number of mole (`n`) and molarity `(C_M)` is: `C_M=n/V`, where `V` is the volume of the solution.

Thus, `n=C_MxxV`.

Therefore, `(C_MxxV)_(H^+)=(C_MxxV)_(OH^(-)`

`=>(C_M)_(OH^-)` `=((C_MxxV)_(H^+))/V=(1.0Mxx28.0cancel(mL))/(20.0cancel(mL))=1.4M`

`=>(C_M)_(NaOH)=1.4M`

Here is a video that might help you with the experimental procedure and calculations for the titration.
Lab Demonstration | Acid - Base Titration.