A) If you make up a solution of 50 mL of 0.1 M TRIS in the acid form, what will be the pH?
(pka of TRIS: 8.3, molecular wt: 121.1 g)
B) if you add 2mL of 1 M NaOH to the solution, what will be the pH?
Thank you.
A) To calculate the pH of the solution, we need to find the hydrogen ion concentration [H+]. Since we have the pKa of TRIS (8.3) and the concentration of TRIS (0.1 M), we can use the Henderson-Hasselbalch equation to determine the pH:
pH = pKa + log ([A-] / [HA])
In this case, the TRIS is in the acidic form, so the ratio of [A-] to [HA] is 1:1; therefore, the log term will be 0.
pH = 8.3 + log (1) = 8.3
So the pH of the 50 mL solution of 0.1 M TRIS in the acid form will be 8.3.
B) To find the pH after adding 2 mL of 1 M NaOH, we first need to determine how many moles of NaOH are added and how many moles of TRIS are present in the solution:
moles of NaOH = (2 mL) * (1 M) = 0.002 moles
moles of TRIS = (50 mL) * (0.1 M) = 005 moles
When NaOH is added it reacts with acidic TRIS (HA) to form the conjugate base (A-) and water, so the reaction is:
HA + NaOH → A- + H2O
Subtracting the moles of NaOH from the moles of HA:
moles of HA remaining = 0.005 moles - 0.002 moles = 0.003 moles
Since 0.002 moles of NaOH react with 0.002 moles of HA, we will have:
moles of A- formed = 0.002 moles
Now we can use the Henderson-Hasselbalch equation again to find the pH:
pH = pKa + log ([A-] / [HA])
pH = 8.3 + log (0.002 / 0.003)
pH = 8.3 + log (2/3)
pH ≈ 8.1
So, the pH after adding 2 mL of 1 M NaOH to the solution will be approximately 8.1.
A) To find the pH of a 0.1 M TRIS solution, we first need to calculate the concentration of TRIS after making up a 50 mL solution.
Concentration (C) = molarity (M) x volume (V)
C = 0.1 M x 50 mL / 1000 mL (conversion to liters)
C = 0.005 moles
Next, we can calculate the moles of TRIS in the acid form by considering the pKa value.
pH = pKa + log[A^-]/[HA]
Given: pKa = 8.3
At equilibrium, [A^-] = [HA]
[A^-] = [HA] = 0.005 mol
pH = 8.3 + log(0.005/0.005)
pH = 8.3 + log(1)
pH = 8.3 + 0
pH = 8.3
Therefore, the pH of the 50 mL 0.1 M TRIS solution in the acid form is 8.3.
B) If you add 2 mL of 1 M NaOH to the solution, we need to consider the reaction between NaOH and TRIS.
NaOH + TRIS (acid form) -> Na+ + TRIS (base form) + H2O
Since NaOH is a strong base and TRIS is a weak acid, we can assume that TRIS would be completely converted to its base form. Therefore, after the reaction, we will have 0.005 moles of TRIS in the base form.
The pH can now be calculated using the concentration of TRIS in the base form.
pH = pKa + log[A^-]/[HA]
[A^-] = 0.005 moles (since all TRIS has converted to base form)
[HA] = 0 (since all TRIS has converted to base form)
pH = 8.3 + log(0.005/0)
pH = 8.3 + log(infinity)
pH = 8.3 + infinity
The pH is undefined since we cannot take the logarithm of infinity. However, it will be significantly higher than the initial pH of 8.3 before adding NaOH, indicating a basic solution.
Please note that this calculation assumes that the volume change upon adding NaOH is negligible.
A) To determine the pH of a solution of 0.1 M TRIS in the acid form, we need to consider the acid-base equilibrium of TRIS.
TRIS can exist in two forms: the acid form (HTRIS) and the base form (TRIS-). In an aqueous solution, these two forms can exchange protons based on the following equilibrium:
HTRIS ⇌ H+ + TRIS-
Given the pKa of TRIS (8.3), we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log [A-]/[HA]
where [A-] is the concentration of the base form (TRIS-) and [HA] is the concentration of the acid form (HTRIS).
In this case, we have 50 mL of 0.1 M TRIS, which means we have 0.1 moles of TRIS in 50 mL. Since the molecular weight of TRIS is 121.1 g, the mass of TRIS in the solution is:
0.1 moles * 121.1 g/mole = 12.11 g
To calculate the concentration of TRIS in the solution, we need to convert the mass to moles:
12.11 g / 121.1 g/mole = 0.1 moles
Since the solution is 50 mL, the concentration of TRIS is:
0.1 moles / 0.05 L = 2 M
The concentration of TRIS- is the same as the concentration of TRIS because the ratio of TRIS- to HTRIS is 1:1. Therefore, [A-] = [HA] = 2 M.
Using the Henderson-Hasselbalch equation:
pH = 8.3 + log [2]/[2]
Simplifying, we find:
pH = 8.3 + log 1
Since the log of 1 is zero, we can conclude that the pH of the solution of 0.1 M TRIS in the acid form is equal to the pKa, which is 8.3.
B) If you add 2 mL of 1 M NaOH to the solution, it will react with the acid form of TRIS (HTRIS) to form water and the base form of TRIS (TRIS-). The reaction can be represented as follows:
HTRIS + NaOH ⇌ TRIS- + H2O
To determine the pH of the resulting solution, we need to consider the new concentration of TRIS- and HTRIS. Since 2 mL of 1 M NaOH is added, the moles of NaOH added can be calculated as:
1 M * 0.002 L = 0.002 moles NaOH
Since the ratio of NaOH to HTRIS is 1:1 in the reaction, the moles of HTRIS that reacts with NaOH is also 0.002 moles.
As we have previously calculated, the initial concentration of TRIS was 2 M. Since the moles of HTRIS that react with NaOH is 0.002 and the ratio of HTRIS to TRIS- is 1:1, the new concentration of TRIS- after the reaction is:
[TRIS-] = [HTRIS reacted] = 0.002 moles / 0.05 L = 0.04 M
To calculate the new concentration of HTRIS, we subtract the concentration of TRIS- from the initial concentration of TRIS:
[HTRIS] = [TRIS] - [TRIS-] = 2 M - 0.04 M = 1.96 M
Using the Henderson-Hasselbalch equation as before:
pH = pKa + log [0.04]/[1.96]
Simplifying, we find:
pH = 8.3 + log (0.04/1.96)
You can solve this equation using a calculator or logarithmic tables to find the pH of the solution after the addition of NaOH.