Question

In the titration of wine to determine the acid concentration, 10.0 mL of wine was placed in a beaker and diluted with 40.0 mL of water. 8.20 mL of 0.051 M NaOH was required to reach the endpoint. Remembering that the acid in wine is tartaric acid, a diprotic acid, what is the molarity (M/L) of tartaric acid in this sample of wine

Answer 1

0.0042 M is the molarity of tartaric acid in this sample of wine.

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 tartaric 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=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`

Putting values in above equation, we get:

`2* M_1* 50.0 mL=1* 0.051 M* 8.20 mL`

`M_1=(1* 0.051 M* 8.20 mL)/(2* 50.0 mL)=0.0042 M`

0.0042 M is the molarity of tartaric acid in this sample of wine.