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

• 0.1255 M

Step-by-step 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

Answer 2

Answer: The concentration of NaOH solution is 0.25 M

Step-by-step 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* 0.1053M* 17.88=1* M_2* 15`

`M_2=0.25`

Therefore, the concentration of NaOH solution is 0.25 M