The endpoint in a titration of a 25 mL sample of aqueous [tex]Ba(OH)_2[/tex] was reached by the addition of 46.5 mL of a 0.1 M [tex]HCl[/tex] titrant. What is the concentration of [tex]Ba(OH)_2[/tex]?

In the titration process of a 25ml Ba(OH)2 sample with 46.5ml of 0.1M HCl, the concentration of Ba(OH)2 is calculated to be 0.093M based on stoichiometry and titration principles.

Explanation:

The concentration of Ba(OH)2 can be calculated using stoichiometry and the principles of titration.

The titration reaction being: Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq).

Given that 46.5ml of 0.1M HCl is needed to neutralize the Ba(OH)2, we notice that the molar ratio between Ba(OH)2 and HCl is 1:2.

Hence, the moles of HCl used are:

(0.1 mol/L) * (46.5/1000 L) = 0.00465 mol.

Following the 1:2 ratio, the moles of Ba(OH)2 are 0.00465/2 mol = 0.002325 mol.

The concentration of Ba(OH)2 is therefore the moles of Ba(OH)2 divided by its volume, so:

(0.002325 mol) / (25/1000 L) = 0.093 mol/L (or 0.093M).

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