Question

How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?

Answer 1

Answer:

$V_{LiOH}=21.8mL$

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

$n_{LiOH}=n_{HBr}$

That it terms of molarities and volumes we have:

$M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}$

Next, solving for the volume of lithium hydroxide we obtain:

$V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL$

Best regards.