Question

A 10.00-ml sample of vinegar, an aqueous solution of acetic acid (hc2h3o2), is titrated with 0.5062 m naoh, and 16.58 ml is required to reach the equivalence point.

a. what is the molarity of the acetic acid?

Answer 1

V(vinegar)*M(viniger)=V(NaOH)*M(NaOH)
10.00 ml*M(viniger)=16.58 ml*0.5062 m
M(viniger, another name of the acetic acid)= 16.58 ml*0.5062 m/10.00 ml= =0.8393m

Answer 2

Answer: The molarity of acetic acid is 0.840 M

Step-by-step explanation:

To calculate the concentration 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 acetic acid

`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=?M\\V_1=10.00mL\\n_2=1\\M_2=0.5062M\\V_2=16.58mL`

Putting values in above equation, we get:

`1* M_1* 10.00=1* 0.5062* 16.58\\\\M_1=(1* 0.5062* 16.58)/(1* 10.00)=0.840M`

Hence, the molarity of acetic acid is 0.840 M