How many milliliters of 0.20N sodium hydroxide must be added to 75 mL of 0.050 N hydrochloric acid to make a neutral solution?

Answer 1

Approx. #19*mL#

We need (i) a stoichiometric equation in order to inform our calculations. Now with one equiv of simple acids and bases, this is trivial:

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

#"Moles of HCl"=75xx10^-3cancelLxx0.050*mol*cancel(L^-1)#

#3.75xx10^-3*mol# with respect to #HCl#.

And thus we need an equiv of #NaOH#, and we have #0.20*mol*L^-1# available, so..........we divide the molar quantity of #HCl# by the #NaOH# concentration to get the req'd volume

#(3.75xx10^-3*mol)/(0.20*mol*L^-1)=0.01875*L=18.8*mL#

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Answer 2

Ellie Andrews

You would need to add 30 milliliters of 0.20N sodium hydroxide to 75 milliliters of 0.050N hydrochloric acid to make a neutral solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.