Answer:
160 mL
Explanation:
Given data:
Molarity of NaOH = 0.105 M
Volume of H₂SO₄ = 40.0 mL (40.0/1000 = 0.04 L)
Molarity of H₂SO₄ = 0.210 M
Volume of NaOH required = ?
Solution:
Number of moles of of H₂SO₄:
Molarity = number of moles / volume in L
0.210 M = number of moles / 0.04 L
Number of moles = 0.210 mol/L × 0.04 L
Number of moles = 0.0084 mol
now we will compare the moles of H₂SO₄ with NaOH.
H₂SO₄ : NaOH
1 : 2
0.0084 : 2/1×0.0084 = 0.0168 mol
Volume of NaoH:
Molarity = number of moles / volume in L
0.105 M = 0.0168 mol / Volume in L
Volume in L = 0.0168 mol / 0.105 M
Volume in L = 0.16 L
Volume in mL:
0.16 L × 1000 mL/ 1L
160 mL
The volume of the NaOH solution required for the reaction is 160 mL
The balanced equation for the reaction is given below:
H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O
From the balanced equation above,
Finally, we shall determine the volume of the base, NaOH required for the reaction.
MaVa / MbVb = nA/nB
(0.21 × 40) / (0.105 × Vb) = 1 / 2
8.4 / (0.105 × Vb) = 1 / 2
Cross multiply
0.105 × Vb = 8.4 × 2
0.105 × Vb = 16.8
Divide both side by 0.105
Vb = 16.8 / 0.105
Vb = 160 mL
Thus, the volume of the base, NaOH required for the reaction is 160 mL
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