How many mL of 0.105 M aqueous NaOH solution are required to neutralize 40.0 mL of 0.210 M aqueous H2SO4 solution? H 2SO 4 + 2 NaOH ---> Na 2SO 4 + 2 H 2O

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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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

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