Question

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?

Answer 1

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