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

`V_(base)=24.04mL`

Step-by-step 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)=(M_(acid)V_(acid))/(2M_(base))`

And we plug in to obtain:

`V_(base)=(0.319M*50.8mL)/(2*0.337M)\\\\V_(base)=24.04mL`

Best regards!