To determine the volume of 0.1 M sodium hydroxide (NaOH) required to neutralize the excess sulfuric acid (H₂SO₄) left after reacting with 6.2 g of NaOH, we'll start by calculating the number of moles of sulfuric acid and sodium hydroxide involved in the reaction.
Step 1: Calculate the moles of sulfuric acid.
- Molarity (M) = Moles of solute/Volume of solution (L)
- Given: 30 cm³ of 1 M sulfuric acid
Convert volume to liters: \[ 30 , \text{cm}^3 = 0.030 , \text{L} \]
Calculate the moles of H₂SO₄: \[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume in L} = 1 , \text{mol/L} \times 0.030 , \text{L} = 0.030 , \text{mol} \]
Step 2: Calculate the moles of sodium hydroxide added.
The molar mass of NaOH (Na = 23, O = 16, H = 1) is: \[ \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 , \text{g/mol} \]
Moles of NaOH: \[ \text{Moles of NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.2 , \text{g}}{40 , \text{g/mol}} = 0.155 , \text{mol} \]
Step 3: Determine the reaction between H₂SO₄ and NaOH.
The balanced equation for the neutralization of sulfuric acid with sodium hydroxide is: \[ \text{H₂SO₄} + 2\text{NaOH} \rightarrow \text{Na₂SO₄} + 2\text{H₂O} \]
From the equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
Calculate the moles of NaOH required to neutralize all the H₂SO₄:
For 0.030 moles of H₂SO₄, the required moles of NaOH: \[ \text{Moles of NaOH required} = 2 \times \text{Moles of H₂SO₄} = 2 \times 0.030 = 0.060 , \text{mol} \]
Step 4: Determine the amount of NaOH in excess.
Since we initially added 0.155 moles of NaOH and only 0.060 moles were required to neutralize the sulfuric acid, the moles of excess NaOH are: \[ \text{Excess NaOH} = \text{Total NaOH} - \text{Required NaOH} = 0.155 , \text{mol} - 0.060 , \text{mol} = 0.095 , \text{mol} \]
Step 5: Calculate the volume of 0.1 M NaOH needed to neutralize the excess acid.
Now, we need to neutralize the excess sulfuric acid that was not neutralized by the sodium hydroxide already present in the solution.
The 0.095 moles of NaOH are needed to neutralize the remaining moles of H₂SO₄.
From the balanced equation: \[ \text{1 mole of H₂SO₄} \text{ requires } 2 \text{ moles of NaOH} \]
If there are 0.030 moles of H₂SO₄ initially and we used 0.060 moles of NaOH to neutralize that, the remaining moles of H₂SO₄ are negligible given the complete neutralization with the added NaOH amount.
Step 6: Calculate moles of 0.1 M NaOH needed.
For every mole of H₂SO₄ remaining, we would need 2 moles of NaOH. If we assume some left over, which is immaterial based on calculations that suggest complete neutralization was achieved.
Conclusion:
The initial neutralization was complete with the accurate amount of NaOH. \[ \text{So, } 0 \text{ extra amount of } 0.1M \text{ NaOH solution is needed given an initial saturation with NaOH.} \]
The calculation is indicating enough initial reactive concentrations have already achieved a neutral balance.
Therefore no volume of 0.1 M NaOH is necessary, all sulfide was taken out with the previous NaOH amount under the given amounts.
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