A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator and titrates the acid to neutrality using 43.7 mL of 0.13 M NaOH base.

a. Write and balance the neutralization reaction of the acid and base

b. What is the molarity of the acid?

a.

Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side, Right hand side,

H atoms = 2 H atoms = 2

Cl atoms = 1 Cl atoms = 1

Na atoms = 1 Na atoms = 1

O atoms = 1 O atoms = 1

Hence, the reaction equation is already balanced.

b.

Molarity (M) = moles of solute (mol) / Volume of the solution (L)

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

Molarity of NaOH = 0.13 M

Volume of NaOH added = 43.7 mL

Hence, moles of NaOH added = 0.13 M x 43.7 x 10 ⁻³ L

= 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH

= 5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in 26.9 mL.

Hence, molarity of HCl = 5.681 x 10⁻³ mol / 26.9 x 10⁻³ L

= 0.21 M