Question

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to you there is a small jar labeled “15.00 g sodium bicarbonate”. Will this be enough sodium bicarbonate to neutralize the spilled sulfuric acid? Show your work and state all reasoning.

Answer 1

First you need a balanced equation that describes the reaction between sulfuric acid and sodium bicarbonate:

H2SO4(aq) + 2NaHCO3(s) --> Na2SO4(aq) + 2H2O(l) + 2CO2(g)

Next you need to calculate how many moles of H2SO4 are present in 85.00 mL of 1.500 M sulfuric acid.

Molarity = moles of solute / liters of solution

1.500 M = n / 0.08500 L

n = 0.1275 mol H2SO4

Now set up and solve a stoichiometric conversion from moles of H2SO4 to grams of NaHCO3. For this you'll need to know the molar mass of NaHCO3.

1 x Na = 1 x 22.99 = 22.99
1 x H = 1 x 1.01 = 1.01
1 x C = 1 x 12.01 = 12.01
3 x O = 3 x 16.00 = 48.00
Total = 84.01 g/mol

0.1275 mol H2SO4 x (2 mol NaHCO3 / 1 mol H2SO4) x (84.01 g NaHCO3 / 1 mol NaHCO3)

= 21.42 g NaHCO3

So unfortunately, 15.00 grams of sodium bicarbonate will *not* be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.