### Educational Content on Acid-Base Titrations#### Problem 4The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution?**Explanation:**In this problem, the titration involves a diprotic acid, meaning each molecule of acid can donate two protons (H⁺ ions). The reaction with NaOH can be represented as:\[ 2 \text{ NaOH} + \text{H}_2\text{A} \rightarrow \text{Na}_2\text{A} + 2 \text{H}_2\text{O} \]Given:- Volume of NaOH used = 21.37 mL- Concentration of NaOH = 0.1432 M- Volume of acid = 10.00 mLThe concentration of the diprotic acid can be calculated using stoichiometry.---#### Problem 510.00 mL of vinegar (mass = 10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the endpoint. Calculate the molarity and mass percent of the acetic acid in the vinegar.**Explanation:**This problem involves determining the concentration of acetic acid (a monoprotic acid) in vinegar. The balanced chemical equation for its reaction with NaOH is:\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]Given:- Volume of NaOH used = 14.77 mL- Concentration of NaOH = 0.4926 M- Volume of vinegar = 10.00 mL- Mass of vinegar = 10.05 gCalculate the molarity of acetic acid and then find its mass percent in the vinegar.---#### Problem 6A 0.2602 g sample of an unknown monoprotic acid requires 12.23 mL of 0.1298 M NaOH solution to reach the endpoint. What is the molecular mass of the acid?**Explanation:**In this scenario, the unknown monoprotic acid

Acid-base titration: Titration is a method to determine the concentration of unknown acid by using the known concentrate base or vice-versa. At the endpoint of the titration, the mole of acid is equal to the mole of the base. The mole of acid or base can be calculated by multiplying of molarity of acid or base by the volume. So, we can say that at the endpoint: For the diprotic acid the above equation is written as follow:For the given titration: Molarity of the acid =? The volume of the acid = 10.00 mL Molarity of the base = 0.1432 M The volume of the base =21.37 mL By rearranging the above equation, we get: Let's put these values into the above formula:The concentration of the diprotic acid is = 0.1530 M Note: According to the guidelines in one session only one question can be solved, please put another question in the next session.