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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?

Sagot :

Sebassandinidn

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!

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