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Question: What volume (in ml) of 0.2 $M$ H$_2$SO$_4$ solution should be mixed with the 40 ml of 0.1 $M$ NaOH s...

What volume (in ml) of 0.2 MM H2_2SO4_4 solution should be mixed with the 40 ml of 0.1 MM NaOH solution such that the resulting solution has the concentration of H2_2SO4_4 as 655M\frac{6}{55}M?

A

70

B

45

C

30

D

58

Answer

70

Explanation

Solution

Let the volume of 0.2 M H2_2SO4_4 solution be VV ml.

  1. Calculate the initial moles of H2_2SO4_4 (in terms of VV) and NaOH. nH2SO4,initial=Molarity×Volume=0.2 mol/L×V×10−3 L=0.2V×10−3 moln_{H_2SO_4, initial} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times V \times 10^{-3} \, \text{L} = 0.2V \times 10^{-3} \, \text{mol}. nNaOH,initial=Molarity×Volume=0.1 mol/L×40×10−3 L=4×10−3 moln_{NaOH, initial} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 40 \times 10^{-3} \, \text{L} = 4 \times 10^{-3} \, \text{mol}.

  2. Write the balanced reaction equation and identify the stoichiometry. H2_2SO4_4(aq) + 2NaOH(aq) →\rightarrow Na2_2SO4_4(aq) + 2H2_2O(l)

  3. Since the final solution contains H2_2SO4_4, NaOH is the limiting reactant.

  4. Calculate the moles of H2_2SO4_4 that react with the limiting amount of NaOH. nH2SO4,reacted=12×nNaOH,initial=12×4×10−3 mol=2×10−3 moln_{H_2SO_4, reacted} = \frac{1}{2} \times n_{NaOH, initial} = \frac{1}{2} \times 4 \times 10^{-3} \, \text{mol} = 2 \times 10^{-3} \, \text{mol}.

  5. Calculate the moles of H2_2SO4_4 remaining by subtracting the reacted moles from the initial moles. nH2SO4,remaining=nH2SO4,initial−nH2SO4,reacted=(0.2V×10−3)−(2×10−3)=(0.2V−2)×10−3 moln_{H_2SO_4, remaining} = n_{H_2SO_4, initial} - n_{H_2SO_4, reacted} = (0.2V \times 10^{-3}) - (2 \times 10^{-3}) = (0.2V - 2) \times 10^{-3} \, \text{mol}.

  6. Calculate the total volume of the mixed solution. Vtotal=V ml+40 ml=(V+40) ml=(V+40)×10−3 LV_{total} = V \, \text{ml} + 40 \, \text{ml} = (V+40) \, \text{ml} = (V+40) \times 10^{-3} \, \text{L}.

  7. Use the given final concentration of H2_2SO4_4, the remaining moles of H2_2SO4_4, and the total volume to set up an equation. Concentration of H2_2SO4_4 = Moles of H2SO4 remainingTotal volume\frac{\text{Moles of H}_2\text{SO}_4\text{ remaining}}{\text{Total volume}} 655 M=(0.2V−2)×10−3 mol(V+40)×10−3 L\frac{6}{55} \, M = \frac{(0.2V - 2) \times 10^{-3} \, \text{mol}}{(V+40) \times 10^{-3} \, \text{L}}

  8. Solve the equation for VV. 655=0.2V−2V+40\frac{6}{55} = \frac{0.2V - 2}{V+40} 6×(V+40)=55×(0.2V−2)6 \times (V+40) = 55 \times (0.2V - 2) 6V+240=11V−1106V + 240 = 11V - 110 240+110=11V−6V240 + 110 = 11V - 6V 350=5V350 = 5V V=3505=70V = \frac{350}{5} = 70.

The volume of 0.2 M H2_2SO4_4 solution required is 70 ml.