Answer:
[tex]V_{base}=24.04mL[/tex]
Explanation:
Hello!
In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:
[tex]2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O[/tex]
We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:
[tex]2M_{base}V_{base}=M_{acid}V_{acid}[/tex]
Therefore, for is to compute the volume of the used base, we proceed as shown below:
[tex]V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}[/tex]
And we plug in to obtain:
[tex]V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL[/tex]
Best regards!
