How many milliliters of 0.250M #NaOH# are required to neutralize 30.4 mL of 0.152 M #HCl#?

1 Answer
May 12, 2016

Approx. #19*mL#. We know that

#"Concentration"="Amount of substance in moles"/"Volume of solution"#

Explanation:

#"Moles of HCl"# #=# #30.4xx10^-3cancelLxx0.152*mol*cancel(L^-1)# #=# #4.62xx10^-3*mol#.

We need an equivalent quantity of sodium hydroxide; #0.250*mol*L^-1# sodium hydroxide is available.

So, #(4.62xx10^-3*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1# #=# #??mL#

What I should have done at the beginning is to write the stoichiometric equation:

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

Why should I write this first? Because it establishes the stoichiometry; the 1:1 molar equivalence. One equiv of sodium hydroxide reacts with one equiv hydrochloric acid. Thus if I know the quantity of acid, I also know the quantity of base. There will be reactions where a different stoichiometry pertains.