A buffer is prepared by mixing 204 mL of 0.452 MHCl and 0.500 L of 0.400 M sodium acetate. (See

A buffer is prepared by mixing 204 mL of 0.452 MHCl and...

Key Concepts

Effect of Adding Strong Base to a Buffer

When a strong base is added to a buffer, it reacts with the weak acid component to form more of its conjugate base. This changes the ratio of conjugate base to weak acid, thereby altering the pH. The buffer's ability to moderate pH changes is quantified by how much strong base or acid it can absorb before the pH shifts significantly.

Mole Calculations in Acid-Base Reactions

Understanding and calculating the number of moles of acids and bases present in solutions is crucial for determining the composition of the buffer. These calculations are often based on the volume and molarity of the solutions involved, and they set the foundation for applying the Henderson-Hasselbalch equation and predicting the pH of the system.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation links the pH of a buffer solution to the acid dissociation constant (pKa) and the ratio of the concentrations of the conjugate base to the weak acid. It provides a convenient way to calculate the pH of a buffer and predict the effects of adding acid or base by relating the change in concentrations of the buffering components.

Buffer Solution

A buffer solution is a mixture that resists changes in its pH when small amounts of acid or base are added. It typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The presence of both components allows the solution to neutralize added acids or bases, thereby stabilizing the pH.

Transcript

00:01 In this problem, we know that our buffer is prepared by mixing 204 milliliters of 0 .452 molar of hcl and 0 .5 liters of 0 .4 molars of sodium acetate.

00:18 We have two parts to this problem.

00:22 A, we need to find the ph of this buffer solution.

00:28 And for b, we need to know how many grams of k -o -h must be added to 0 .5 liters of the buffer to change the ph by 0 .15 units.

00:42 So in this problem, we are given this information.

01:13 We will first need to solve for part a.

01:17 We can first find the walls of hcl.

01:26 204 milliliters is equivalent to 0 .204 liters a mole we know that one molar is a mole over a liter so we can just multiply the volume by its molarity so we know that there are 0 .092 moles of hcl next, we need to find the moles of sodium acetate.

02:17 We do not need to do any extra conversion because it is already in liters, and we can directly do the multiplication to find the number of moles.

02:44 If we compare these two values, we can see that the number of moles for hcl is less than the number of moles for sodium acetate.

02:54 So we know that hcl is the limiting reagent.

02:59 It will run out first.

03:04 Given this information, we can establish an ice table.

03:10 We can first write our equation of the reactions and our ice table, which stands for initial change and equilibrium.

03:50 0 .200 and 0.

03:53 Because we are trying to find the ph of the buffer solution, these two are what we are considering.

04:04 We do not need to consider sodium chloride.

04:10 Our change, we are going to use up all of our hcls because that is our limiting reagent.

04:22 And at equilibrium, here are our values.

04:31 And our total volume, as we know, we have used 204 milliliters of hcl and 0 .5 liters of sodium acetate.

04:47 All we need to do is do the addition.

04:54 We will need to use this information to find the number of moles for the acid and the conjugate base.

05:02 So we will calculate this portion.

05:08 We know that one molarity is a mole over a liter.

05:14 So given what we know already from our calculations, given over here, and our final volume, we will do the same for the conjugate base, which gives us zero, 1 .131 molar.

06:07 Next, we will need to use the henderson hasselback equation, which states that the p .k., the ph equals to the pka plus the log of the acid over the conjugate base.

06:33 And to solve this, we need to know the k .a value of acetic acid, which is 1 .8 times 10 to the negative 5.

06:43 We know that the p -k -a is the negative log of our k -a, and so using our given information from the appendix, we know that the p -k -a is 4 .74.

07:01 We can plug in the rest of our information into the equation to solve for the ph.

07:08 Our ph is 4 .81.

07:25 That is our solution to part a.

07:29 For part b, we can find, since we are trying to find how many grams of k -oh must be added to the 0 .5 liters of buffer to change the ph by 0 .15 units, we can first see, solve for the change.

07:57 So our ph is 4 .81...

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