Which of the following would form a buffer if added to 650.0...

Which of the following would form a buffer if added to 650.0 mL of $0.40 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_2$ ?
(a) 1.00 mol of HF
(b) 0.75 mol of HF
(c) 0.30 mol of HP
(d) 0.30 mol of NaP
(e) 0.30 mol of HCl

Key Concepts

Buffer Solutions

Buffer solutions are mixtures that resist changes in pH when small amounts of acid or base are added. They are typically made up of a weak acid and its conjugate base, or a weak base and its conjugate acid, which work together to neutralize added acids or bases.

Acid?Base Neutralization

Acid?base neutralization involves the reaction between an acid and a base. In the context of buffer formation, a weak acid reacting partially with a strong base (or vice versa) can produce a solution containing appreciable amounts of both a weak acid and its conjugate base, which is necessary for a buffer to function.

Stoichiometry in Buffer Formation

Understanding stoichiometry is essential when mixing solutions to form a buffer. The mole ratio of the weak acid to the strong base (or weak base to strong acid) determines whether there will be sufficient amounts of both the conjugate acid and conjugate base in the final solution, which is critical for maintaining the buffer capacity.

Role of Strong and Weak Electrolytes

The distinction between strong electrolytes, which completely dissociate in solution, and weak electrolytes, which only partially dissociate, is important in buffer systems. The complete dissociation of strong electrolytes provides a predictable source of ions, while the partial dissociation of weak acids or bases allows for the dynamic equilibrium necessary for the buffer action.

Transcript

00:01 So we're going to try to figure out which of these things are going to be buffers.

00:05 So let's start by finding out what we know about our strontium hydroxide.

00:12 So i'm going to take the molarity times of liters and see that i have 0 .26 moles of strontium hydroxide.

00:21 We're going to multiply that by 2 and see that we have 0 .52 moles of oh minus.

00:28 So these things are going to be added to that much oh minus.

00:34 So in our first example, we're adding some hf that's going to react with our oh minus.

00:42 It's going to make some h2o and some f minus.

00:47 I'm going to do a little table, initial change, and final.

00:52 I start with one mole here and 0 .52.

00:57 I don't care about the water.

00:58 Obviously the oh minus is our limiting reactant.

01:02 So minus 0 .52 of all these things plus 0 .52 here.

01:06 So i end up with 0 .48 moles of my weak acid and 0 .52 moles of my weak base.

01:18 So since i have a weak acid and it's conjugate weak base, this is a buffer.

01:28 And our second example, we're simply going to do the same thing.

01:31 We're just going to use less oh minus and see what happens.

01:35 Okay, so same i, c, final...

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