Can someone PLEASE help me with this problem!?
Here's the problem I'm working on:
A buffer that contains .11 M HY (conjugate acid) and .220 M Y- (conjugate base) has a pH of 8.77. What is the pH after .001 mol Ba(OH)2 is added to .75L of solution?
[[HY and Y- are standard forms of a conjugate acid/base pair, its not a specific buffer]].
Answer:
[HY] = 0.11 M
[Y(-)] = 0.22 M
pH = 8.77
Okay. You plug these values into your buffer equation and solve to get the pKa of HY. That's just plug-and-chug.
When you add Ba(OH)2, you're doing this:
2 HY + Ba(OH)2 = BaY2 + 2 H2O
So, 0.001 mol of Ba(OH)2 neutralizes 0.002 mol of HY and forms 0.002 mol of Y(-). Here we are assuming that BaY2 is soluble so that it will exist as ions in solution.
Your volume is 0.75 L, and you have 0.11 moles per liter of HY, so that's 0.0825 mol of HY present. After adding the barium hydroxide, you have 0.0805 mol of HY in 0.75 L, which corresponds to a concentration of 0.1073 M.
You have 0.22 moles per liter of Y(-), so that's 0.1650 moles of Y(-) present. Your neutralization reaction created 0.002 mol of Y(-), so now you have 0.1670 moles of Y(-), in 0.75 L, which corresponds to a concentration of 0.2227 M.
[HY] = 0.1073
[Y(-)] = 0.2227 M
pKa = obtained in first step above
Plug into buffer equation and get the new pH.
Victoria,
I don't normally respond to questoins like this one that actually have a numerical problem to work out. But since no one responded to your first posting, and you seem to genuinely want to understand the problem, I will outline for you how to do the problem.
The pH of a buffer depends on just two things: the Ka of the conjuate acid, and the mole ratio between the base species and the acid species. This is expressed in the Henderson-Hasselbach equation:
pH = pKa + log [Y-]/[HY]
where pKa is the negative logarithm of the acid's dissociation constant (analogous to the formula for pH, just substitute Ka for [H+]). In your problem, you are given the pH of the initial buffer, and the concentrations of the base and acid. Simply plug these values into the formula above, and you can solve for the pKa of the acid.
Now that you have this information, you can solve the rest of the problem. It tells you that you add base to the solution... this base is going to react with the acid part of your buffer and consume some of the HY. You need to find out how much will be consumed. This means you have to set up a soichiometry problem. Write out the equation for this reaction, it should look like this:
2HY + Ba(OH)2 ----> 2H2O + BaY2
You must now use this equation to convert the moles of Ba(OH)2 given in the problem to the amount of HY consumed. If 0.001mol of Ba(OH)2 reacted, how many moles of HY reacted? How many moles of HY were there to begin with? (Hint: use the original molarity together with the volume) How many moles of HY are left?
Now that you know how many moles of HY remain in the solution, you can divide this by .75 L to get the new concentration of HY. Finally, your goal is to calculate the new pH of the solution. To find the pH of a buffer, remember that we just need to use the Henderson-Hasselbach equation. So set up that equation again... you already solved for pKa in the first step, so you can put in this value right away. You have just calculated the new molarity of acid from the paragraph above, so plug this is for [HY], and the molarity of the base species can stay the same. Simply plug and chug, and out comes the new pH.
I have not worked out the numerical solution partly because I don't have a calculator handy, and partly because I don't like the idea of simply handing the answer over to students on a silver platter (I taught chemistry for 4 years). But I hope this explanation helps, and gives you what you need to solve the problem on your own. If you are still really confused, you can e-mail me with questions.
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