Question

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

Approx. `20*mL`.......

Explanation:

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.