What volume of 0.475 M hydrochloric acid solution reacts with 36.4 mL of 0.350 M sodium hydroxide solution?

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Answer 1 · 2017-05-06T04:24:38

We interrogate the stoichiometric reaction...........and get a volume of approx. #27*mL# of the given acid.

We interrogate the stoichiometric reaction...........

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

The reaction has 1:1 stoichiometry, and thus moles of acid are equivalent to moles of base.

Now #"concentration"="moles of solute"/"volume of solution"#;

And thus #"moles of solute"="concentration"xx"volume of solution"#.

And thus.....................................

#n_"NaOH"=36.4xx10^-3*Lxx0.350*mol*L^-1=1.27xx10^-2*mol#

And given the molar quantity, we may this work out the required volume of #HCl#:

#"Volume"_"HCl"=(1.27xx10^-2*mol)/(0.475*mol*L^-1)xx1000*mL*L^-1=#

#=26.8*mL#........