A.) A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. If the titration required 17.73 mL of 0.1036 M NaOH to reach the equivalence point, calculate the concentration (in M) of the weak acid in the sample. B.) If the sample solution described in A). contained 0.1845 g of the Weak Acid, calculate the molar mass (in g/mol) of the weak acid.

A) 0.1225 M

B) 100.4 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

17.73 mL of 0.1036 M NaOH react. The reacting moles are:

0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.

Step 4: Calculate the molar concentration of HA

1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:

M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M

Step 5: Calculate the molar mass of HA

1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:

0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.