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3) 10 mL of a 0.1 M NaOH solution were titrated against a 0.05 M HCl solution. How much (in mL) of the 0.05 M HCl solution will be required to reach the end-point of the titration?

Answer 1

Olivia Anton

Approx. #20*mL#.......

We need to establish the stoichiometry....

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

And so 1:1 stoichiometry operates.

We use the relationship, #"concentration"="moles of solute"/"volume of solution"#

And so #"moles"="volume"xx"concentration"#

With respect to #NaOH#, we have a molar quantity of ...

#10*mLxx10^-3*L*mL^-1xx0.10*mol*L^-1=1.0xx10^-3*mol.#

And thus, there will be a ............

#(1.0xx10^-3*mol)/(0.05*mol*L^-1)xx10^3*mL*L^-1=20*mL# volume required for equivalence.

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Answer 2

Cameron Andrews

5 mL of the 0.05 M HCl solution will be required to reach the end-point of the titration.

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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.