How many mL of a 3.2M NaOH solution is required to neutralize 74mL of a 12.4M HCl solution?

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Answer 1 · 2016-05-05T07:57:02

Approx. 300 mL

For all these sorts of problems we require a balanced chemical equation:

#NaOH(aq) + HCl(aq) rarr NaOH(aq) + H_2O(l)#

Such a scheme establishes that there is a 1:1 equivalence between moles acid and moles of base.

#"Moles of hydrochloric acid"# #=# #74xx10^-3Lxx12.4*mol*L^-1=0.895*mol#.

Thus we require a quantity of #0.895*mol*"NaOH"#:

#(0.895*cancel(mol))/(3.2*cancel(mol)*L^-1)# #=# #0.28*L" sodium hydroxide solution"#

This quantity is reasonable as the the acid is 4 times as concentrated as the base.