Question

Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalence point?

Answer 1

Answer:

$V_{base}=24.04mL$

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

$2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O$

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

$2M_{base}V_{base}=M_{acid}V_{acid}$

Therefore, for is to compute the volume of the used base, we proceed as shown below:

$V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}$

And we plug in to obtain:

$V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL$

Best regards!