Refer to the balanced chemical equation:
2 NaHCO3 (s) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O (l) + 2 CO2 (g)
If you spilled sulfuric acid on the benchtop and solid sodium bicarbonate is added until the fizzing, caused by the formation of carbon dioxide, stops. If 35 mL of 5.5 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?
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From the balanced chemical equation, we can see that the molar ratio between H2SO4 and NaHCO3 is 1:2. That means for every 1 mole of H2SO4, we need 2 moles of NaHCO3 to neutralize it.
First, we need to calculate the number of moles of H2SO4 spilled.
Molarity (M) is defined as moles of solute per liter of solution. We are given the volume of H2SO4 (35 mL) and its concentration (5.5 M). To convert mL to L, we divide by 1000.
Volume of H2SO4 = 35 mL = 35/1000 L = 0.035 L
Now we use the formula for calculating moles:
Moles of H2SO4 = concentration * volume in liters
= 5.5 M * 0.035 L
= 0.1925 moles
Next, we use the molar ratio to determine the moles of NaHCO3 needed.
Moles of NaHCO3 = moles of H2SO4 * (2 moles NaHCO3 / 1 mole H2SO4)
= 0.1925 moles * (2 moles NaHCO3 / 1 mole H2SO4)
= 0.385 moles
Finally, we can calculate the mass of NaHCO3 needed using its molar mass.
The molar mass of NaHCO3 is:
Na: 1 * 22.99 g/mol = 22.99 g/mol
H: 1 * 1.01 g/mol = 1.01 g/mol
C: 1 * 12.01 g/mol = 12.01 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
Total molar mass = 22.99 + 1.01 + 12.01 + 48.00 = 83.01 g/mol
Mass of NaHCO3 = moles * molar mass
= 0.385 moles * 83.01 g/mol
= 31.96 g
Therefore, the minimum mass of NaHCO3 that must be added to neutralize the acid is 31.96 grams.
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