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

Question

Consider the titration of a 25.0-mL sample of 0.115 M RbOH 
with 0.100 M HCl. Determine each quantity. 
c. the pH at 5.0 mL of added acid

Accepted Answer

Hello everyone today. We are being asked to examine the following questions. So it says, what is the ph of a solution of 25 mL of 250.239 moller of this cesium hydroxide. That is tight. Trade with five mL of 50.283 molars of H N. 03 or nitric oxide. So the first thing I'm gonna do is we want to find the number of moles of each of our Compounds. So what we do is we take our volume, which is 25 ml. We convert this to leaders because In terms of polarity it's over leaders and so we can say that we have 1000 middle leaders per one leader and we simply can multiply by the number of moles that we have. So we're going to do .239 mold of cesium hydroxide. This is going to give us 5.9 Eight times 10 to the negative 3rd moles of cesium hydroxide. And we're gonna do the same thing with the nitric oxide. So we're going to take five ml, we're going to convert that to leaders. We're gonna multiply by the number of moles that we have. So we have .283 moles Of nitric oxide. And when we do that we're going to get 1. Times 10 to the negative 3rd moles of nitric oxide. Next we're gonna set up our ice table here. So we're gonna have cesium hydroxide at the first column we're gonna say plus nitric oxide is going to yield us water and cesium N. 03. And so we're gonna plug in the values that we have for these. So for season hydroxide initial we have 5.98 times 10 to the negative third For nitric oxide. We have 1.42 times 10 to the negative third. And then of course water is always going to be zero across the board. And then for our C. S. N. 03 we're going to have zero as well. Next we're going to write what the least denominator is in this situation. So that's going to be 1.415 or 1.42 times 10 to the negative third. And so we're gonna write that for both of these values here I'm gonna say 1.42 times 10 to the negative third. Of course this is going to give us zero for nitric oxide But this value is going to give us 4.56 times 10 to the negative third for C. S. O. H. Next we're also going to say that when we have that the ph is dependent on our excess base integrations. And so we're going to have a total volume from both of these compounds of 25 mL plus five middle leaders to give us 30 middle leaders And we convert this to leaders that gives us 0. leaders. And so to find the concentration of hydroxide ions. We're going to take O. H. We're going to take our 4.56 times 10 to the negative third. We're going to divide that by the number of leaders that we have. That's going to give us 0.15 to moller. But since we're asked to find the ph we now have to do P. O. H. Of this would be negative log of the concentration, which is 0.152. That would give us 0.8182. And of course ph is 14 minus the p. O. H. And so we're going to take 14 and subtract it by 0.8182 to get a final ph of 13.18 as our final answer, I hope this helped. And until next time.

Consider the titration of a 25.0-mL sample of 0.115 M RbOH with 0.100 M HCl. Determine each quantity. c. the pH at 5.0 mL of added acid

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