Respuesta :
Ans: Volume of NaOH needed = 50.3 ml
Given:
Molarity of NaOH = 0.200 M
Molarity of H2SO4 = 0.479 M
Volume of H2SO4 = 10.5 ml
To determine:
Volume of NaOH required
Explanation:
The titration reaction can be shown as:
2NaOH + H2SO4 → Na2SO4 + 2H2O
Based on the reaction stoichiometry:
2 moles of NaOH reacts with 1 mole of H2SO4
i.e.
moles of NaOH required = 2(moles of H2SO4)
[tex]V(NaOH) * M(NaOH) = 2[V(H2SO4) * M(H2SO4)]\\\\\\V(NaOH) = \frac{2[10.5 ml*0.479M}{0.200M} \\ = 50.3 ml[/tex]
The volume of 0.200 M NaOH needed to required to titrate 10.5 mL of 0.479 M H2SO4 is 50.3 mL
We'll begin by calculating the number of mole of H₂SO₄ in the solution. This can be obtained as follow:
Volume of H₂SO₄ = 10.5 mL = 10.5 / 1000 = 0.0105 L
Molarity of H₂SO₄ = 0.479 M
Mole of H₂SO₄ =?
Mole = Molarity × Volume
Mole of H₂SO₄ = 0.479 × 0.0105
Mole of H₂SO₄ = 5.0295×10¯³ mole
Next, we shall determine the number of mole of NaOH needed to react with 5.0295×10¯³ mole of H₂SO₄. This can be obtained as follow:
H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O
From the balanced equation above,
1 mole of H₂SO₄ reacted with 2 moles of NaOH.
Therefore,
5.0295×10¯³ mole of H₂SO₄ will react with = 5.0295×10¯³ × 2 = 1.0059×10¯² mole of NaOH.
Finally, we shall determine the volume of the NaOH solution needed.
Mole of NaOH = 1.0059×10¯² mole
Molarity of NaOH = 0.2 M
Volume of NaOH =?
Volume = mole / Molarity
Volume of NaOH = 1.0059×10¯² / 0.2
Volume of NaOH = 0.050295 L
Multiply by 1000 to express in mL
Volume of NaOH = 0.050295 × 1000
Volume of NaOH = 50.3 mL
Therefore, the volume of NaOH needed for the reaction is 50.3 mL
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