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

a polar bond is formed when electrons are shared unevenly between atoms