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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?

Respuesta :

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: The molarity of acetic acid is 0.840 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is acetic acid

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.5062M\\V_2=16.58mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 10.00=1\times 0.5062\times 16.58\\\\M_1=\frac{1\times 0.5062\times 16.58}{1\times 10.00}=0.840M[/tex]

Hence, the molarity of acetic acid is 0.840 M

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