Suppose you add 2.0 mL of 0.10 M HCl to 100 ml of buffer having 0.10 M HA and 0.20 M NaA.
a. Which species will react strongly with one another?
I believe the answer is Cl- and Na+ to form NaCl.
b. What will be the change in pH? pKa = 4.82
I am not sure how to find the answer to this one.
a is not right.
You are adding an acid to a buffer. The base will be the one reacting with the acid and the base is A^-.
b. In chemistry the first two thoughts about working a problem are as follows:
1. mols
2. ICE table
In a buffer solutiion, I usually do these by millimoles and not by M (concentrations).
mmols HA = 100 mL x 0.1 M = 10
mmols A^- = 100 mL x 0.2 M = 20
mmols HCl added = 2 mL x 0.1M = 0.2
.......A^- + H^+ ==> HA
I.....20.....0.......10
add.........0.2................
C....-0.2..-0.2.....+0.2
E.....19.8...0.......10.2
Substitute the E line into the HH equation and calculate pH then take the difference from the initial pH to find the change.
I got pH = 6.76
a. Well, it seems you've got the right idea! When you mix HCl and NaA, the Cl- ions from HCl will react with the Na+ ions to form NaCl. So, indeed, Cl- and Na+ will strongly react with each other!
b. Ah, the change in pH, a puzzly question! To find the change in pH after the addition of HCl, we need to think about what happens to the acid and its conjugate base in the buffer solution.
Since we're adding HCl, the HA in the buffer will react with it, forming A-. This reaction will consume some of the HA, reducing its concentration. And as the concentration of HA decreases, the concentration of its conjugate base, A-, will increase.
To calculate the change in pH, we need to know the initial concentrations of HA and A-. Could you provide those values? Then we can crunch some numbers and have a pH party! 🎉
To answer part b, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the acid and its conjugate base. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pH is the desired pH, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, we are adding a strong acid (HCl) to a buffer solution containing HA and NaA. The strong acid will react with the conjugate base (A-) to form the weak acid (HA).
Initially, the concentrations are:
[HA] = 0.10 M
[A-] = 0.20 M
After the reaction, a portion of the [A-] will be converted to [HA]. Let's say x moles of A- react. This means that x moles of HA will be formed. The final concentrations can be calculated using the stoichiometry of the reaction:
[HA] = 0.10 M + x
[A-] = 0.20 M - x
Since x is small compared to the initial concentrations, we can assume that 0.20 M - x ≈ 0.20 M and 0.10 M + x ≈ 0.10 M.
Plugging these values into the Henderson-Hasselbalch equation:
pH = 4.82 + log(0.20 / 0.10)
pH = 4.82 + log(2)
Using logarithm properties, log(2) = 0.3010:
pH = 4.82 + 0.3010
pH ≈ 5.1210
So the change in pH after adding the HCl to the buffer solution will be approximately 5.1210 - 4.82 = 0.3010.
To determine the change in pH when adding an acid to a buffer solution, we need to consider the Henderson-Hasselbalch equation, which describes the relationship between the pH of a buffer solution, the pKa of the acid, and the ratio of the concentrations of the acid to its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where:
- pH is the measure of acidity or alkalinity
- pKa is the logarithmic constant of the acid dissociation constant (Ka)
- [A-] is the concentration of the conjugate base (in this case NaA)
- [HA] is the concentration of the acid (in this case HA)
In our case, the initial concentrations of HA and A- are:
[HA] = 0.10 M
[A-] = 0.20 M
When we add 2.0 mL of 0.10 M HCl, it will react with NaA to form NaCl, as you mentioned. However, the amount of HCl added is small compared to the total volume of the buffer solution. Therefore, we can neglect the change in concentration of NaA and assume that it remains the same.
The key is to understand that when we add HCl, it reacts with the buffer's conjugate base (NaA) to form the weak acid (HA). So the change in concentration of HA will be due to the reaction with HCl.
To find the change in concentration of HA, we can use the dilution formula:
C1V1 = C2V2
where:
- C1 and V1 represent the initial concentration and volume of the acid (HCl)
- C2 and V2 represent the final concentration and volume of the acid (HCl)
In our case, C1 = 0.10 M (the concentration of HCl) and V1 = 2.0 mL (the volume of HCl added).
C2 represents the new concentration of HA after the reaction, and V2 = 100 mL (the total volume of the buffer solution).
By rearranging the formula, we can solve for C2:
C2 = (C1V1) / V2
= (0.10 M * 2.0 mL) / 100 mL
= 0.002 M
So the final concentration of HA is 0.002 M.
Now, we can substitute the values into the Henderson-Hasselbalch equation to find the change in pH:
pH = pKa + log([A-]/[HA])
= 4.82 + log(0.20 M / 0.002 M)
= 4.82 + log(100)
= 4.82 + 2
= 6.82
Therefore, the change in pH when adding 2.0 mL of 0.10 M HCl to the buffer solution will be an increase of 2 units, from the initial pH of the buffer.