A buffer that contains 0.40 M of a base, B, and 0.25 M of its conjugate acid, BH^+, has a pH of

A buffer that contains 0.40 M of a base, B, and 0.25 M of...

Key Concepts

Acid-Base Neutralization in Buffers

When a strong acid like HCl is added to a buffer, it reacts with the basic component (the weak base) of the buffer to form more of its conjugate acid. This reaction alters the ratio between the base and its conjugate acid, which, according to the Henderson-Hasselbalch equation, results in a change in pH. Understanding this neutralization process is key to predicting how the buffer pH will shift.

Buffer Solutions

Buffers are solutions that resist changes in pH when small amounts of acid or base are added. They are typically composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. This property is important because it maintains a relatively constant pH, which is crucial in many chemical and biological systems.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a simple way to estimate the pH of a buffer by relating the pH to the pKa (or pKb) and the ratio of the concentrations of the conjugate base to its conjugate acid. This equation is fundamental when analyzing how the buffer pH changes as a result of adding an acid or base.

Stoichiometric Calculations in Buffers

Stoichiometric calculations are used to determine the number of moles of the buffering components before and after the reaction with an added acid or base. By calculating the changes in the moles of the weak base and its conjugate acid, one can update their concentrations in the solution. These adjusted concentrations are then used in the Henderson-Hasselbalch equation to determine the new pH of the buffer.

Transcript

00:01 So normally when we have a weak acid conjugate base system, we apply the henderson -hasobach equation.

00:09 But in this case, we have the opposite of that, and we have a base and a weak base in a conjugate acid.

00:16 So we have to apply essentially the similar equation for conjugate bases and conjugate acids and weak bases.

00:30 So we're given information that the ph of this weak -base.

00:36 Conjugate acid solution is 8 .88 and the poh is 14 minus the ph which means that the poh for this is 5 .12 and we're given information that our initial concentration of our conjugate acid is equivalent to 0 .25 molars.

00:57 Our initial concentration of our weak base is equivalent to 0 .40 molars.

01:08 So now we can essentially use this to find the kb for the weak base.

01:14 This becomes 0 .25 over 0 .40.

01:20 So we can plug in and solve for the pkb in this case.

01:32 And we find that essentially the kb in this case is equivalent to 4 .8 times 10 to the negative 6.

01:42 And this piece of information will be useful later.

01:44 Now we're given information that we're going to stress the system by adding hcl.

01:52 So hcl in this case only can react with the base since the conchid acids is already protonated and a double protonation wouldn't be favorable.

02:03 So here we have our base reacting with h -plus leads to, we can assume this is a complete reaction since there's a significant difference in the relative of the two compounds.

02:17 So this leads to the formation of bh plus.

02:21 So through this process, the concentration of base will decrease, while the concentration of our conjugate acid will increase.

02:31 So we're given information that the volume of our solution is 0 .25 liters.

02:38 And since we know that concentration is as expressed as the number of moles over volume, we can find that the number of moles of base that we have initially is equivalent to 0 .4 times 0 .25, which is equivalent to 0 .10 moles.

02:55 And by adding 0 .002 0 .4 of strong acid, this would be subtracted and become, so minus this quantity here, this would become 0 .0080 moles...

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