Consider the titration of a 35.0-mL sample of 0.175 M HBr with
0....

Question

Consider the titration of a 35.0-mL sample of 0.175 M HBr with
0.200 M KOH. Determine each quantity.
e. the pH after adding 5.0 mL of base beyond the equivalence
point

Accepted Answer

Hello everyone today. We are being given the following problem and asked to solve for it. So it says after the addition of five mL of base beyond equivalence point, what is the ph of the solution of 15 mL of 150.126 Mueller nitric acid treated with 0.145 molar sodium hydroxide. So the first thing that we want to do is we want to find our moles of nitric acid. And to do that we take our 15 middle leaders. We convert this into leaders by saying that one leader is equal to 1000 mL. And then we finally multiplied by the malaria E, which is 0.126 moles per one leader. Our units will cancel out and we'll be left with 1.89 times 10 to the negative 3rd moles of nitric acid. Next we want to calculate our volume of our base. And by doing that we can take the moles that we just calculated for the 1. times 10 to the negative third moles of our nitric acid. We're gonna go ahead and multiply by the multiple ratio that we have. So for every one mole of nitric acid We use one roll of sodium hydroxide to titrate it begin, then multiply by our molar itty. And by that we have one leader per 0.145 moles of sodium hydroxide. Our units will cancel And we will be left with the value of 0.01303 leaders of our sodium hydroxide. However, we want this value to be in terms of milliliters. And so to do that, we can simply use the conversion factor that there are 1000 mL per one leader giving us after our units cancel out of course 0.3 Mila leaders. Step three involves finding our volume, our total volume from which we just calculated here. And that's going to be our 15 Metal leaders plus our 13.03 ml And then plus our five middle leaders, giving us 33.03 middle leaders. And for all intents and purposes, we're going to convert this into leaders by saying that we have ml for one leader. And when our units cancel, We're left with 0. leaders. Just a couple more steps, we're going to go ahead and find how many moles of our base we have. So we calculated the volume. Now we want to find the moles of our base. And so what we're gonna do is we're gonna take our five middle leaders of our base that we added, We're going to convert that into leaders by saying that one leader is equal to middle leaders. And then we're going to calculate by our polarity of our base which is going to be 0.145 moles Of sodium hydroxide, our base over one leader. And when our units cancel, We're going to be left with 7.25 times 10 to the negative 4th moles of our base or sodium hydroxide. And so ultimately we want to find the ph and so we already have our concentration or we can already solve our concentration of hydroxide ions by using our moles of our base 7.25 times 10 to the negative fourth moles Over our volume, which is 0.03303 l To get 0.02195 as our polarity. We then can solve our P. O. H. With the equation that the P O. H. Is equal to r negative log of our concentration of hydroxide ions. And that's simply going to be negative log of 0.2195 or 1.6 586. Our last and final step step seven over here is going to be the ph plus our P. O. H. Is equal to 14 to find the P. O. H. Or the ph We simply just plug in our P. O. H. Here 1.6586 equals 14. And solving for P. H. We get a final value of 12.34. I hope this helped until next time

Consider the titration of a 35.0-mL sample of 0.175 M HBr with 0.200 M KOH. Determine each quantity. e. the pH after adding 5.0 mL of base beyond the equivalence point

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