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

rArr Normality=2*Molarity

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),

`88*2.6*2=V*1.6`

`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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