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

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:

$n_1M_1V_1=n_2M_2V_2$

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:

$2\times M_1\times 50.0 mL=1\times 0.051 M\times 8.20 mL$

$M_1=\frac{1\times 0.051 M\times 8.20 mL}{2\times 50.0 mL}=0.0042 M$

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