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

Chemistry

1 answer:

slavikrds [6]2 years ago

8 0

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.

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the first three ionization energies of an element x are 590, 1145, and 4912 kj mol-1. what is the most likely formula of the com

joja [24]

Answer: The chemical formula is $XSO_3$

Explanation:

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The ionization energy equation for the given values follow:

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From the values of ionization energy above, it can be seen that the ionization energy increases as every succeeding electron is removed.

Second ionization energy is a little higher than the first one but there is a huge amount of difference between the third and second ionization energy.

This implies that the ion formed during second ionization energy has a stable configuration and it requires a humongous amount of heat to release the third electron.

Hence, the ion formed will be $X^{2+}$