A buffer that contains 1.05M B and 0.750M BH^+ has a pH of 9.50 . What is the pH after 0.0050 mo

step 1 So to answer this question, I've put our pH up here as 9 .5. And the concentration of the base w

A buffer that contains 1.05M B and 0.750M BH^+ has a pH of...

A buffer that contains 1.05$M \mathrm{B}$ and 0.750$M \mathrm{BH}^{+}$ has a $\mathrm{pH}$ of $9.50 .$ What is the pH after 0.0050 $\mathrm{mol}$ of $\mathrm{HCl}$ is added to 0.500 $\mathrm{L}$ of this solution?

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Key Concepts

Acid-Base Stoichiometry in Buffers

In buffer problems, stoichiometry is used to account for the reaction between added strong acids or bases and the corresponding buffer component. By calculating the moles of buffer components before and after the reaction, one can determine the new ratio of conjugate base to acid, which in turn is used to recalculate the pH using the Henderson-Hasselbalch equation.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH of a buffer to the pKa (or pKb) and the ratio of the concentrations of the conjugate base and acid. It is a fundamental tool in calculating the pH of buffer solutions, especially when the buffer has been perturbed by the addition of a strong acid or base.

Buffer Solutions

Buffer solutions are mixtures of a weak acid and its conjugate base or a weak base and its conjugate acid, which resist significant changes in pH upon the addition of small amounts of strong acid or strong base. This stability is due to the equilibrium between the acid and base components that can absorb added H+ or OH- ions.

Transcript

00:01 So to answer this question, i've put our ph up here as 9 .5.

00:06 And the concentration of the base we find is 1 .05 molar.

00:12 The concentration of the acid, its conjugate acid, is 0 .75 molar.

00:17 And the volume we are working with is 0 .5 liters.

00:20 So i put up the henderson -hasselbach equation up here for us to reference because the first step is to find the pca.

00:29 And so if you rearrange the equation, it will be pga equals ph minus the log of the concentration of the base over the concentration of the acid.

00:54 Okay.

00:57 So we just plug in the numbers.

01:00 Pca equals 9 .5 minus log of 1 .05 molar over 1 .05.

01:09 0 .75 molar.

01:12 Okay, and you plug that in your calculator and you'll find that the pga equals 9 .35.

01:20 Okay.

01:21 And we'll need to remember that for the future because we will use that again.

01:27 So the next step here is to find the concentration of the hydronium ions in solution or the the hydrogens, the positive hydrogens that the hydrochloric acid puts into the solution.

01:45 And the way you do that is by just taking moles over the volume, which equals molarity.

01:51 It's pretty straightforward there.

01:53 So we added 0 .005 moles of hydrochloric acid to 0 .5 liters.

02:06 Oh, that's a weird 5.

02:07 And that equals 0 .01 molar.

02:12 Okay...

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