Question

What is the molarity of an acetic acid solution if 25.0 mL of 0.212 M NaOH solution is required to neutralize 16.5 mL of the HC2H3O2 solution?

HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq)

Answer 1

The molarity of an acetic acid solution : 0.321 M

Further explanation

Titration is a procedure for determining the concentration of a solution by reacting with another solution that is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Reaction

HC₂H₃O₂(aq) + NaOH(aq) → H₂O(l) + NaC₂H₃O₂(aq)

Molarity NaOh=0.212

volume NaOH=25 ml=0.025 L

mol NaOH =

`\tt mol=M* V\\\\mol=0.212* 0.025\\\\mol=0.0053`

mol ratio NaOH : HC₂H₃O₂ from the equation = 1 : 1, so mol HC₂H₃O₂ :

`\tt (1)/(1)* 0.0053=0.0053`

The molarity of acetic acid HC₂H₃O₂ :

`\tt molarity(M)=(n)/(V)\\\\M=(0.0053)/(0.0165)\\\\M=0.321`

Or you can use titration formula :

`\tt M_1V_1n_1=M_2V_2n_2(n=acid/base~valence, for NaOH~and~Acetic~acid=1)\\\\0.212* 25* 1=M_2* 16.5* 1\\\\M_2=(0.212* 25* 1)/(16.5* 1)\\\\M_2=0.321`