An auto mechanic spills 88 mL of 2.6 M H2SO4 solution from a rebuilt auto battery. How many milliliters of 1.6 M NaHCO3 must be poured on the spill to react completely with the sulfuric acid?
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Acid-base neutralization reactions can be traced on the basis of equivalents.In other words, 1 equivalent of an acid reacts completely with and neutralizes 1 equivalent of a base.
Thus, for a neutralization reaction,
`V_1*N_1=V_2*N_2` --- (i)
(Where, V-terms refer to volume and N-terms, Normality)
H2SO4 is a dibasic acid, so its 1 mole= 2 equivalents.
Hence, in its solution, Molarity = Normality/2
NaHCO3 is a monoacidic base, so its 1 mole= 1 equivalent.
So, in its solution, Molarity = Normality
Let the volume of 1.6 M NaHCO3 solution required for complete neutralization be V mL.
Plugging in the values in eqn. (i),
`rArr V=(88*2.6*2)/1.6=286` mL.
Therefore, 286 mL of NaHCO3 solution will be required to react completely with the spilled sulfuric acid.
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