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Ch.17 - Aqueous Ionic Equilibrium

Chapter 17, Problem 72a

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. Calculate the pH at each volume of added base: 0 mL.

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Hey everyone we're asked what is the ph of 45 mL of 0.187 moller economic asset tight traded with zero mL of 0.200 molar sodium hydroxide first. Let's go ahead and write out our reaction. We know that economic acid is a weak acid so we can simply write H. A. To represent ethanol like acid. And we react this with water and we end up with the an ion of economic acid plus our hydro ni um ions. We can then create our ice chart. And we know that initially we had 0.187 moller of our economic acid and we had zero of our products. Initially we can disregard our water since it is a liquid and our change is going to be minus X. For our economic acid and a plus X. On our product side. Since we're losing reactant and gaining products in our equilibrium, we have 0.187 minus eggs for our economic acid and an X. And an X. In our products. When we check our textbooks, we find that the K. A. Of economic acid is 1.8 times 10 to the negative five to sell for X. We know that our K. Is equivalent to our products over our reactant. So it would be X. Times X over 0.187 minus X. Now in order to check if we can disregard our X. We can go ahead And take our 0. And divide this by our K. A. of 1.8 times 10 to the -5. We find that our answer comes up to 10,388 which is greater than 500. And since it is greater than 500 we can go ahead and disregard our X. In our denominator. To simplify this a bit further, we have 1.8 times 10 to negative five is going to be equal to X squared over 0.187. We can go ahead and multiply both sides by 0.187. And we then get X squared equals 3.366 times 10 to the negative six. Taking the square root. We end up with X equals 1.8347 times 10 to the -3. And this is going to be the concentration of our hydro ni um ions. And since our question was asking us for the Ph We can then take the negative log of 1.8347 times 10 to the negative third. And this will get us to a ph of 2.74 which is going to be our final answer. So I hope that made sense. And let us know if you have any questions