Acetic Acid Buffer Workup Worksheet
In order to complete the homework for the Acetic Acid Buffer Lecture, you will need to know (remember) some basics we have already covered or that you should have covered in a previous chemistry course.
There are several aspects of acids and bases that you should already know from taking general chemistry 1: Strong versus Weak acids and bases, pH and pOH, and how to calculate the concentration of H+ and OH− from pH and pOH respectively. For this workbook we are just going to practice those concepts you should already know.
Strong versus Weak
What can change the pH of water? Weak electrolytes (salts that form acids or bases in reaction with water) and Acids and Bases are the predominant compounds that change the pH of water.
Weak electrolytes react with water to form acidic or basic solutions:
Al(NO3)3(s) + H2O <=> Al3+(aq) + 3NO3−(aq)
All3+(aq) + H2O(l) <=> Al(OH)3(~aq) + H+(aq) Solution becomes acidic
Strong acids and bases fully dissociate in water:
HCl(aq) → H(aq) + Cl−(aq)
Ca(OH)2(aq) → Ca2+(aq) + 2OH−(aq)
You should know these:
Strong electrolytes like highly soluble salts also fully dissociate in water. Slightly soluble salts and weak acids and bases only partially dissociate in water causing changes in the pH. These are equilibrium events and the concentration of H+ or OH− must be calculated using equilibrium methods (eg. ice tables).
Based on the definition of the “p” in pH as the “power” of hydrogen, the pH is the negative logarithm of the [H+]. The pH value determines the acidity of a solution. As the pH of pure water is 7, this is set as the “neutral” pH with the scale up or down from that point being considered basic (>7) or acidic (<7).
The pH scale is logarithmic which means that each change of a value of 1 unit is actually a 10x change in the concentration of [H+].
pH and pOH Relationship
Another way the concentration of an acid or base solution can be indicated is through pH or pOH. The small p in front of the H or OH is an indicator for a mathematical process called a logarithm. The p can be translated as “take the negative log of” whatever value follows it. In the case of pH and pOH this translates mathematically to:
pH = -log[H+] and pOH = -log[OH–]
In this way, if we know the pH of a solution, we know the concentration of H+ in that solution and conversely, if we know the proton concentration we can calculate the pH of the solution.
[H+] = 10-pH and [OH-] = 10-pOH
If the concentration of H+ is 1.50 x 10−6 M, the pH is -log(1.50 x 10−6 M) = 5.823908 and with correct significant figures this is 5.824.
If the pH of the solution is 8.45, the [H+] = 10−8.45 = 4.0 x 10−9 M
In the second example you might think that the [H+] is very small, but this makes sense because the pH indicates that the solution is basic and not acidic meaning there are very few protons in the solution.
The logarithmic scale allows the small concentrations of protons in the solution to be displayed in easy to manage values. The pH scale indicates which compounds can be defined as acids and which compounds can be defined as bases.
pH values < 7.00 are considered acidic
pH values > 7.00 are considered basic
pH value = 7.00 is considered neutral.
This scale is based on the dissociation of H+ and OH- ions in water.
In pure water the concentration of H+ ions from the dissociation of the water molecules themselves is 1.00 x 10−7 M. This means that the pH value of pure water is 7.00 and has come to mean the point of neutrality in the scale. This is because if the concentration of H+ in pure water is 1.00 x 10-7M then the concentration of OH- must also be 1.00 x 10−7 M as well:
H+ + OH− ⇔ H2O and [H+] = [OH−] = Neutral
Some other information we can derive from this statement for pure water:
pH + pOH = pKw
pH + pOH = -log(1.00 x 10−7 M) + -log(1.00 x 10−7 M) = pKw = -log(1.01 x 10−14) ≈14
-log (1.01 x 10−14) actually = 13.99568
Thus if we know the pH of a solution, we can calculate the pOH by subtracting the pH value from 13.99568 and vice versa.
What is the pH of a solution that has a [OH-] = 2.34 x 10−7M?
- First we need to calculate pOH: pOH = -log(2.34 x 10-7M) = 6.630784
- Second, we subtract the pOH from 14 to get the pH: pH = 13.99568 – 6.630784 = 7.364895 rounding to the correct significant figures the answer is 7.364.
Logarithm Rules for Significant Figures
When you take the Log or Ln of a number X.XX then the answer to that calculation must reflect the number of significant figures in the value within the mantissa (digits after the decimal) of the answer .XXX. So for example, if you take the Logarithm of a number 4.967 x 10−4 M then the answer should have 4 digits after the decimal or 3.3039.
For an antilog (10^ or e^) the process is reversed. This means that the number of digits after the decimal in the number you are taking an antilog for will determine the total number of digits in the answer you should give. For example, e-3.5 has one digit after the decimal so the proper answer should be 3 x 10−2 M.
Polyprotic Acids and Bases
Polyprotic acids and bases are those that release more than one proton or hydroxide ion respectively when dissolved in water. This feature is very important when you are trying to calculate the pH of the solution. For instance, the strong acid H2SO4 (sulfuric acid) is diprotic. This means that when it is dissolved in water it releases 2 protons into the solution. To calculate the pH of this solution, you would need to take the overall concentration of the sulfuric acid and multiply its concentration by 2 prior to calculating its pH.
If you have a 0.250M concentration of H2SO4, then the H+ concentration would be twice the concentration of the overall acid or 2 x 0.250M = 0.500M H+
The pH of the solution would then be calculated as pH = -log(0.500M) = 0.301 (Remember your log rules for sigfigs here. Since there are 3 digits in the value we calculated with there must be 3 digits in the mantissa (amount after the decimal) of the answer)
NOTE: In general chemistry 1 we made the assumption that all acids and bases fully dissociated i.e. acted like a strong acid or base. We now know this is not accurate and thus only the pH of the strong acids and bases in the table above can be calculated directly in this manner. All others must first have their dissociation determined by an equilibrium calculation.
pH Problem #1:
Calculate the [H+] and pH in the following solutions at 25°C, and classify each as acidic , basic or neutral.
pH Problem #2:
pH Problem #3: