A buffer is prepared by mixing 50.0 mL of 0.050 M sodium bicarbonate and 10.7 mL of 0.10 M NaOH

A buffer is prepared by mixing 50.0 mL of 0.050 M sodium...

Key Concepts

Buffer Solution

A buffer solution is a system that resists significant changes in pH upon the addition of small amounts of acid or base. It usually consists of a weak acid and its conjugate base (or a weak base and its conjugate acid), and its ability to stabilize pH is due to the equilibrium established between these components.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simplified formula used to relate the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of its conjugate base to the acid. It is expressed as pH = pKa + log([base]/[acid]), making it a fundamental tool for calculating buffer pH and predicting how changes in component proportions will affect it.

Acid-Base Reaction Stoichiometry

In preparing or modifying a buffer, stoichiometry plays a critical role by governing the reaction between the buffer components and any added strong acids or bases. Understanding the molar relationships in these reactions allows one to determine how additions will shift the balance between the weak acid and its conjugate base, thereby altering the pH of the system.

Buffer Modification and pH Adjustment

Adjusting the pH of a buffer involves adding a measured amount of a strong acid or base, which reacts with one of the buffer components. The change in the ratio of conjugate base to acid is then used in the Henderson-Hasselbalch equation to determine the new pH. This concept is crucial for quantifying how much of an acid or base is needed to achieve a specified change in pH.

Transcript

00:01 Question 30 is an acid -based question regarding the preparation of a buffer and the modification of that buffer in order to achieve a particular ph.

00:15 Typically when working with buffer solutions, we are going to use the henderson -hasselbulch equation in order to solve for ph.

00:24 This question is in reference to a buffer that was prepared by adding 50 milliliters of 0 .050 molar sodium bicarbonate.

00:33 And 10 .7 milliliters of 0 .1 molar sodium hydroxide.

00:39 So there is no buffer solution with just sodium bicarbonate.

00:44 When we add the sodium hydroxide, the sodium bicarbonate will behave as an acid and turn into carbonate.

00:54 Therefore, the base is carbonate and the acid is sodium bicarbonate.

01:04 So when we set up our henderson -hasselbalch equation, ph will be equal to p -ka, or the negative log, of the ca value of the acid.

01:15 We identified the acid as being sodium bicarbonate.

01:20 Going to the appendix, we see that the second ka value for carbonic acid, which corresponds to bicarbonate, is 4 .7 times 10 to the negative 11.

01:34 So, henderson -hasselbalch equation, ph, equals p -k -a, plus the log of, you can either have the base concentration over the acid concentration, or the moles of base over the moles of acid.

01:49 I'm going to use moles.

01:51 So we added 50 .0 milliliters of 0 .5 molar sodium bicarbonate, and 10 .7 milliliters of 0 .1 molar sodium hydroxide.

02:06 Every mole of sodium hydroxide we add will convert one mole of sodium bicarbonate into carbonate.

02:15 So the moles of weak base formed, carbonate, will be equal to the moles of strong base added.

02:22 So the moles of carbonate will be equal to the volume of sodium hydroxide, 10 milliliters or 0 .017 liters multiplied by its concentration, 0 .1 molar.

02:37 The moles of sodium bicarbonate, the acid, will be equal to the moles that we start with.

02:44 Moles can be calculated, just as we did appear, by taking the volume, 50 milliliters or 0 .05 liters, multiplied by the concentration, 0 .05 molar.

02:54 This is the moles of the acid, sodium bicarbonate, that we have before adding the sodium hydroxide.

03:01 When we add the sodium hydroxide, the moles of sodium bicarbonate will decrease by the number of moles of strong base added, and the moles of strong base added will be calculated just as we did up here in the numerator.

03:15 Then what's left is our calculation.

03:17 So when we calculate taking the negative log of this value and the log of this ratio, and we sum them up, we get 10 .20 as our ph.

03:30 So that's just part a.

03:33 Part b is actually the more challenging part of this question.

03:37 How many grams of hcl must be added to 25 milliliters of this buffer to achieve a ph change of 0 .07 units? well, first we need to identify, is this ph change going to be an increase in ph or a decrease in ph? we're adding a strong acid hcl, and a strong acid will result in a decrease in ph.

04:02 Adding a base to a solution increases ph, adding an acid decreases ph.

04:09 So we are going to decrease the ph from 10 .27 by 0 .07 units...

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