Consider the titration of a 25.0-mL sample of 0.115 M RbOH
with 0...

Question

Consider the titration of a 25.0-mL sample of 0.115 M RbOH
with 0.100 M HCl. Determine each quantity.
b. the volume of added acid required to reach the equivalence
point

Accepted Answer

Hello. Everyone in this video we're trying to calculate for the volume of the acid required to reach equivalence point of hydration. So let's recall that at the equivalence point of a tight creation that are moles of acid is going to equal to the moles of our base. And again to recall we have that the polarity is actually equal to our moles over volume. Now switching this equation up and just isolating our moles. We get that. The moles is equal to um polarity times volume. I'm just kind of piecing up this puzzle here. We're gonna go ahead and take this and this and saying that our M. V. Which is just like old times, polarity times volume of our base is going to equal to the malaria times volume of our acid right now. Just scrolling down a little bit. Alright, so we're trying to find the volume of our acid and that's going to equal to the malaria times volume of our base, divided by the clarity of our acid. I'm going to head to plug in some numerical values. Now with this equation We'll have 45.0 ml times are 0.18 Moller as you call moller just most times are over volume. That's why we can go like this and that's going to be divided by R 0.100 molar in that case. Then we'll get the volume of our acid Equaling to again explain this into my calculator. I'll get the value of 81.0 million liters. So the amount of volume that's required to reach the equivalent point of registration is going to equal to 81.0 ml. And that's going to be my final answer for the problem. Thank you all so much for watching.

Consider the titration of a 25.0-mL sample of 0.115 M RbOH with 0.100 M HCl. Determine each quantity. b. the volume of added acid required to reach the equivalence point

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