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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?

User Asok Buzz
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1 Answer

12 votes

Answer:


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!

User Sjaustirni
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