# URGENT Concentration of Vinegar | Titration? [closed]

Example of acid base titration

The moles of acid and base are equal in the end point, so $$n_A=n_B$$ where $$A$$ and $$B$$ are acid and base, respectively. The number of moles can be written as $$n=C*V$$ where $$C$$ is the molar concentration and $$V$$ is the volume. So

$$n_A=n_B$$

$$C_A*V_A=C_B*V_B$$

$$C_A*10mL=15.7 mL*0.05 M$$

Threfore

$$C_A= 0.0785M$$

Now, this is moles of acetic acid per liter. The mass per mole of acetic acid is $$M_r=60,052 g/mol$$. So in grams per liter this is

$$0.0785M*60.052 g/mol= 4.71 g/L$$

So now you have 4.71 g in 1000 mL (1L). The only thing left to do is a simple rule of three because %m/V es mass of acetic acid in 100 mL, not 1000 mL. Then

$$100*4.71/1000=0.471$$

• Welcome to Chem SE! Thank you for your contribution. But this site works on a few principles, we don't answer a question where the OP shows no effort from his side. We don't do the entire problem for them, we only clear conceptual doubts in their workings. We guide them through constructive comments. So I would encourage you to do the same :) Feb 15 '20 at 16:50
• Thank you @Matthew_Adams Feb 16 '20 at 9:22