Respuesta :
Answer:
0.0042 M is the molarity of tartaric acid in this sample of wine.
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 tartaric 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=2\\M_1=?\\V_1=10.0+40.0 mL=50.0 mL\\n_2=1\\M_2=0.051 M M\\V_2=8.20 mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 50.0 mL=1\times 0.051 M\times 8.20 mL[/tex]
[tex]M_1=\frac{1\times 0.051 M\times 8.20 mL}{2\times 50.0 mL}=0.0042 M[/tex]
0.0042 M is the molarity of tartaric acid in this sample of wine.
