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?

1 Answer
Nov 23, 2015

Answer:

#=>(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.