Determine the pH of an HF solution of each concentration. In
whic...

Question

Determine the pH of an HF solution of each concentration. In
which cases can you not make the simplifying assumption that
x is small? (Ka for HF is 6.8 * 10-4.)
a. 0.250 M b. 0.0500 M c. 0.0250 M

Accepted Answer

Hey everyone, we're asked to calculate the ph of 0.300 moller and 0.30 moller of nitrous acid solution. In which case do you have to use the quadratic formula. And they provided us the K. Of nitrous acid. So first let's go ahead and look at our first scenario Where we have 0.300 moller of nitrous acid. First. We want to go ahead and create our ice chart. So we have our nitrous acid and this is going to react with water. Now since we have a weak acid, we're going to produce our nitrite ion plus our hydro ni um ion Creating our ice chart. We know that initially we had 0.300 moller of our nitrous acid. And we're going to disregard our water in our expression since it is a liquid and we had zero of our products. Initially our change is going to be a minus X. On our react inside and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium on our react inside, we have 0.300 minus X. And an X. And an X. In our product side. Now let's go ahead and look at our K. A. Which we know to be the concentration of our products over our reactant. So this will be the concentration of our hydro ni um ion times the concentration of our nitrite ion all divided by the concentration of our nitrous acid. So plugging in these values, we know that our K. Is 4.0 times 10 to the negative four. And this is going to be equal to X times X divided by 0.300 minus X. Now we can go ahead and check if the X. And our denominator is negligible By taking that value of 0.300 and dividing it by RK. A. of 4.0 times 10 to the -4. Now, if we get a value greater than 500 then we can safely disregard the X. In our denominator. And in this case we actually do get a value greater than 500 so we can disregard this. X. Simplifying this a bit further, we have 4.0 times 10 to the negative four is equal to X squared over 0.300 Multiplying both sides by 0. and subsequently taking the square root of both sides. We end up with an x. 0.01095 moller which is going to be the concentration of our hydro ni um ion. So to calculate our ph we can go ahead and take the negative log of the concentration of our hydro ni um ion. So plugging in that value, we have the negative log of 0.1095 and this will get us to a ph of 1.96 which is going to be our answer for our scenario one. Now let's go ahead and look at scenario two Where we have 0.030 moller of nitrous acid. Again, we're first going to create that ice chart where we have nitrous acid and this will react with water. When this reacts with water, we get our nitrite ion plus our hydro ni um ion creating our ice chart. We know that initially we had 0.030 moller of our nitrous acid and we're going to disregard our water in our expression since it is a liquid and we have zero of our products formed. Initially our change is going to be a minus X on our react inside and a plus X on our product side. Since we're losing reactant and gaining products equilibrium in our react inside, we have 0.30 minus X. And an X. And an X in our product side. So we're going to take the same steps as we did in the previous scenario where we know that our K. A. Which is equal to 4.0 times 10 to the negative four. And this is equal to our products over our reactant. So in this case we have X times and X all over 0. minus X. Again, we want to check if the X in our denominator is negligible and we can do so by taking 0.30 and dividing this by our K. A. Of 4.0 times 10 to the negative four. If we get a value greater than 500 then we can safely omit the x in our denominator. But in this case we get a value less than 500. So we have to keep the X. In our denominator. So simplifying this further, we have 4.0 times 10 to the negative four is equal to X squared over 0.30 minus X. So now we're going to move this around so we can use our quadratic formula. So we get X squared plus 4.0 times 10 to the negative four X -1. times 10 to the -5. Which is going to be equal to zero. Now let's go ahead and use our quadratic formula to solve for X. So we get X is equal to -4.0 times 10 to the - plus or minus the square root of four point oh times 10. To the negative four squared minus four times one times negative, 1.2 times 10 to the negative five. And this will be divided by two times 1. Now when we calculate for x we get two values. Our first value is going to be negative 3.66987 times 10 to the negative three. Now our second value comes up to x equals 3. Times 10 to the -3. Now our value cannot be negative. So this means that the correct value is going to be our second value. So to calculate our ph we're going to take the negative log of the concentration of our hydro ni um ion, which is our X. And in this case it is 3. Times 10 to the -3 Moller. And this gets us to a ph of 2.49. So to answer this question, it looks like Case two or scenario two requires the quadratic formula. Now I hope this made sense and let us know if you have any questions.

Determine the pH of an HF solution of each concentration. In which cases can you not make the simplifying assumption that x is small? (Ka for HF is 6.8 * 10-4.) a. 0.250 M b. 0.0500 M c. 0.0250 M

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