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Calculate the pH at the following points in a titration of 40 $\mathrm{mL}(0.040 \mathrm{L})$ of 0.100 $\mathrm{M}$ barbituric acid $\left(K_{\mathrm{a}}=9.8\right.$ $\times 10^{-5} )$ with 0.100$M \mathrm{KOH}$$$\begin{array}{l}{\text { (a) no KOH added }} \\ {\text { (b) } 20 \mathrm{mL} \text { of KOH solution added }} \\ {\text { (c) } 39 \mathrm{mL} \text { of KOH solution added }} \\ {\text { (d) } 40 \mathrm{mL} \text { of KOH solution added }} \\ {\text { (e) } 41 \mathrm{mL} \text { of KOH solution added }}\end{array}$$

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Chemistry 102

Chapter 14

Acid-Base Equilibria

Liquids

Drexel University

University of Maryland - University College

University of Kentucky

Brown University

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Let's go with the pH of the following points of a tight tray. Shin, we have 40 mL of 0.100 Moeller barbecue Eric acid, which has the formula C four each four and 23 So for a we're going to calculate, uh, de ph with no kill each added. Yeah. So weak acid, Whether equilibrium. Here c four h three n 203 minus And we're told that we have 0.1 Moeller pleader ice table here. Que a sequel to you h 30 c. Four h three and 203 minus over C four h four in 203 and were given the day is 9.8 attempts 10 to the negative. Five is equal to x X one minus x. We do get a quadratic equation here, Solve for X. I'm gonna plug this into my equations over. So Okay, 98 times 10 to the negative five. Okay, if you need God's X and X in the numerator and 0.1 minus Excellent. The denominator solving we find will take the positive route. Too sick Pigs gives me 3.1 times 10 to the negative three h 30 plus with a 3.1 times 10 to the negative. Three. Mueller the pH vehicle to negative log 3.1 times 10 to the negative three, and we'll get a PH. Equal to 251 So that would be the pH for part a with no k o. H. Added for Part B. We're going to add 20 mL of K o. H. Just being added, and we need to find the pH. So we're gonna calculate the moles of O. H. Minus. We're told that the mill arat e of the potassium hydroxide is 0.1. Moeller, 20 mL is 0 to 00 leaders, and this would be equal to 0.200 bulls, of which minus, we'll find the most of the acid. He four h four and 203 And it is 0.1 Moeller and 40 mL of the acid 0400 leaders. And this is equal to 0.400 moles of the acid C four, h four in 203 And now we'll set up equation here where we've got the acid C four age four into three the O. H. Minus on this would be C four h three, n 203 minus on H 20 We're going to complete this size table in malls and most of the acid 0.400 Moles of the base. His 0.200 zero neutralized the acid and we'll be left with 00200000200 Andi, Now we can use the Henderson Hasselbach equation. The P H is equal of peak a plus. The log of the conjugal piece over Yeah, acid P H is equal to the peak A for the negative log of K 9.8 times 10 to the negative five plus the log of the beasts 0.200 Look up the acid on this be equal to the pH. Equal to 4.1 So upon adding 20 mL of potassium hydroxide, the PH works out to 4.1 For part C, we're gonna add 39 millimeters of the potassium hydroxide. So how to meet the moles of the O. H. Minus just 0.1 times 0.390 leaders. This is going to be equal to 00390 moles of the H minus from part be. We've already calculated the moles of the acid, and that's equal to 0.400 moles of the acid. So using this, let's set up a rice table c four, h four and to hold three at O H minus. Going to see four. H three and 203 minus and each too old on again or his table will be completed in moles 0.400 and 0.390 zero minus 00.390 minus 0.390 plus 0.390 will get +00390 here zero here and 000100 Supplier. Henderson Hasselbach Equation pH is equal to the peak A which is negative log of the A 9.8 times 10 to the negative five. Plus the log. The clarity of base over the majority of the acid on this would be equal to a pH of 5.61 so this would be the ph when, right, the when 39 mL of the husband added for D, we're gonna add 40 milliliters of the coach. So let's calculate the moles of O H minus, which will be 0.1 times 0.0 400 leaders and this workout 2.400 moles of the alleged minus and from parts B, B and C, the moles of the C four each four and 23 is equal to 00400 Moles of the Assad. Yeah. Yeah. So in this case here, yeah, theme with okay. PH of the solution depends on the and I in Okay. Yeah. Sil me Moment. Here. So here this is the pH at the equivalence point where the malls of the acid is equal to the malls of the base. So the ph depends on the C four h four her age three and 203 minus. So the morality of the sea for H three n 203 minus Z equals 2.0 400 moles and total volume is 40 millimeters and 40 millimeters, or 400.800 liters on this will work out 2.500 Mueller c four h three n 203 minus at H 20 Would you will see four in each four in 203 and O H minus. She'll change equilibrium. This is 0.50000 It's complete a rice table 0500 minus x X and X on. This will be based on the baby. What should be see for age four and 23 times O. H minus. C four, h three and 203 minus. And let's find out the KB here before we continue. Which we K W over K a. One point no times 10 to the negative 14. And the is equal to 9.8 times 10 to the negative five. So K B is equal to 1.2 times 10 to the negative. 10. Let's bring that over here. Okay? Substitute our values in and we do get a quadratic so we'll go ahead and solve for X plug that into my occasions over here. Uh huh. Taking the positive roots, I get 2.3 times 10 to the negative six. Well, let's go Let's go. 366226 Yeah, Mhm times 10 to the negative. Six Going back to the ice table, you h minus. It's equal to 2.26 times 10 to the negative. Six smaller You could find the p o H. Do you think of the negative log 2.26 times 10 to the negative six, and that would be equal to 5.65 Ph. 14 minus 5.65 gives us a pH of 8.35 So upon adding 40 million liters of Coke pH, the Ph is equal to 8.35 and last thing here for Part E, we're gonna add 41 mL of coach. So let's first find the moles of the H minus, which is 0.1 at this 0.41 two leaders, which was 00410 moles of O. H. Minus on from the previous questions, the mole C four h four and 203 is equal 2.0 with 0400 malls. Yeah, and we will have some neutralization taking place here. Okay, so the moles of O H minus in excess will be 00.410 moles minus the 0.400 moles. And this will yield, um 0.10 Moles, the final volume sequel, the 40 mL of the asset plus 41 mL of the base. And that's 81 mL. Therefore, the morality of the O. H minus in the final mixture here is pointed reserves or 10 moles, divided by 0.810 leaders. This works out to a molar ity of okay, 0.1 To Mueller. It's one of the pH, which is the negative log 0.12 This will give us a pH of 2.92 PH is 14 minus 292 and this will be called pH of 11.1. So, upon adding 41 mL of K O. H, the pH of the resulting mixture is 11.1

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