A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. (Ka = 3.5 x 10^-4)
(a) 4.11
(b) 3.82
(c) 3.46
(d) 3.09

Question

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. (Ka = 3.5 x 10^-4) 
(a) 4.11
(b) 3.82
(c) 3.46
(d) 3.09

Accepted Answer

hi everyone for this problem it reads find the ph of a two liter solution of 20.11 moller nitrous acid and 0. molar of sodium nitrite upon the addition of 100 and mL of 500.225 molar hydrochloric acid. Note that the P. K. A. Of nitrous acid is 3.25. So the question that we want to answer here is P. H. And then the problem, we're told the concentration of nitrous acid, we're told the concentration of sodium nitrite and we're told the concentration of hydra bronek acid were also given volume of hydroponic acid. Okay, so we need to find P. H. Okay? And we're going to first calculate the moles of nitrous acid, its conjugate base and hydro bronek acid. Okay, so let's go ahead and do that first. So for our moles of nitrous acid, we're going to start with what we're given or we're going to start with the volume. Okay, so we know we have two leaders of solution and we know what the malaria T. Is. So we can use the malaria T given for the nitrous acid to find the moles. So the malaria is 0.11 molar which is moles per liter. So this then when we write it out is 0. moles of nitrous acid per liter of solution. So our units for leaders cancel and we're left with moles. So the molds of nitrous acid is 0. moles. So let's go ahead and do the same thing for moles of N. 02 which is the conjugate base. Okay, and that is coming from here from the sodium nitrite. Okay, so our starting point is going to be the same. We have two leaders of solution and the mill arat, e is the same which is 0. moles her leader. So leaders cancel and we have the same number of moles for the conjugate base. Now let's go ahead and solve for the moles of Hydra Bronek acid. Okay, so we're told that the volume however, for this is 0.15, excuse me, it's 150 ml. This should that should be leaders. Okay, so this should be leaders. So we have 100 and 50 leaders. And the mill arat, e is 0.225 moles Per leader. Okay, so our moles is 0. moles for moles of hydrochloric acid. So upon addition of a strong acid which is our hydro bronek acid. The conjugate base of our nitrous acid reacts to neutralize it. So what that reaction is going to look like is we have our conjugate base is going to react with the strong acid or hydrochloric acid and this is going to produce nitrous acid plus the conjugate base. Okay, so like I said, upon addition of this strong acid, the conjugate base reacts to neutralize it. What we can do here is we can create an ice table so that we can know what the concentrations are at equilibrium. So for our ice table we need to write out ice and I like to draw a line to separate our products from our react mints. We have no solids or liquids in our reaction. So we need to write out the initial concept, we can write out everything here. So let's start off with our initial concentration for our conjugate base, we just saw for that and that value is 0. moles. Okay, for our concentration of hydra bronek acid, It is 0. Molar. And for the concentration of the conjugate base or excuse me for the concentration of The nitrous acid, it is 0.220. Okay, so let's go ahead and do our change in concentration and we're going to subtract Here the Mnolds of Hydra Bronek acid. Okay, so we're going to subtract 0. from the reactant side. And we're going to add 0.03375 to the product side. So for the equilibrium row then we're going to have no hydro bronek acid. And then for our conjugate base we're going to have 0.18625 moles. And then for the nitrous acid we're going to have 0.25375. Now that we know these values, we can use the Henderson Hasselbach equation to determine the ph of the buffer after the addition of a small amount of strong acid. And the Henderson Hasselbach equation is P H. Is equal to P. K. A. Plus the log of the concentration of conjugate base over the concentration of the weak acid. And remember in the problem where we're told what the P. K. Is and now we've solved for the equilibrium concentrations of conjugate base and the weak acid. So we just need to plug everything in. Now let's replace our conjugate base in conjugate acid with the actual conjugate base in conjugate acid. So when we rewrite this, it becomes P K. A. R. P H. Is equal to P K. A. Plus the log of the conjugate base over the con the concentration of the weak acid. Okay, so P H. Is equal to the P. K. A. Or yes, so ph is equal to the P. K. We're told is 3.25. And this is going to be plus the log of the concentration of the Conjugate Base. Look at your ice table and see what you wrote in the equilibrium row, it is 0.18625. And this is over the concentration of the weak acid. 0.25375. Okay, so what we're going to get then for our final answer for PH is p is going to equal 3.12. And this is our final answer. That is it for this problem. I hope this was helpful.

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. (Ka = 3.5 x 10^-4) (a) 4.11 (b) 3.82 (c) 3.46 (d) 3.09

Verified Solution

Key Concepts

Buffer Solutions

Henderson-Hasselbalch Equation

Acid-Base Neutralization