Question

A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equivalence point in a titration. What is the concentration of the naoh solution?

Answer 1

Answer: The concentration of NaOH solution is 0.25 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $H_2SO_4$ solution = 0.1053 M

$V_1$ = volume of $H_2SO_4$ solution = 17.88 ml

$M_2$ = molarity of $NaOH$ solution = 0.46 M

$V_2$ = volume of $NaOH$ solution = 15.0 ml

$n_1$ = valency of $H_2SO_4$ = 2

$n_2$ = valency of $NaOH$ = 1

$2\times 0.1053M\times 17.88=1\times M_2\times 15$

$M_2=0.25$

Therefore, the concentration of  NaOH solution is 0.25 M

Answer 2

Answer:

• 0.1255 M

Explanation:

1) Data:

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

2) Chemical reaction:

The titration is an acid-base (neutralization) reaction to yield a salt and water:

• Acid + Base → Salt + Water

• H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

3) Balanced chemical equation:

• H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

4) Stoichiometric mole ratio:

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

• 1 mole H₂SO₄ : 2 mole NaOH

5) Calculations:

a) Molarity formula: M = n / V (in liter)

   ⇒ n = M × V

b) Nunber of moles of acid:

• nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

• nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

• nₐ = nb

• 0.1053 × (17.88 / 1,000) =  Mb × (15.00 / 1,000)

• Mb = 0.1053 × 17.88  / 15.00 = 0.1255 mole/liter = 0.1255 M