A) If you make up a solution of 50 mL of 0.1 M TRIS in the acid form, what will be the pH?
B) If you add 2 mL of 1M NaOH to the solution in part A, what will be the pH?
If it is needed, here is some addition information: TRIS pka= 8.3
mw of TRIS-HCL= 157.6 g/mol;
mw TRIS (Trizma base)= 121.1 g/mol
Thank you!
To find the pH of the solution in part A, you need to consider the dissociation of TRIS. TRIS can exist in two forms: TRIS-H+ (acid form) and TRIS^- (base form). The pKa value of TRIS is given as 8.3, which represents the pH at which half of the TRIS is in the acid form and half is in the base form.
We can assume that the solution is dilute, allowing us to neglect the contribution of TRIS-H+ to the pH. Therefore, we solely need to consider the concentration of TRIS^-.
First, we need to determine the moles of TRIS in the solution. The concentration of TRIS is given as 0.1 M, so in 50 mL (0.05 L) of solution, there are 0.1 x 0.05 = 0.005 moles of TRIS.
Since the molecular weight (mw) of TRIS is 121.1 g/mol, the mass of TRIS in the solution is 0.005 x 121.1 = 0.6055 grams.
Next, we need to calculate the molar concentration (molarity) of TRIS in the solution. To do this, we divide the moles of TRIS by the volume of the solution in liters:
Molarity (M) = moles/volume (L)
Molarity = 0.005 moles / 0.05 L = 0.1 M
Now that we have the molar concentration of TRIS, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log[base]/[acid]
Given that the pKa value of TRIS is 8.3, and the ratio of [base] (TRIS^-) to [acid] (TRIS-H+) is 1:1 at the pKa, the equation simplifies to:
pH = pKa + log(1)
pH = 8.3
Hence, the pH of the solution in part A is 8.3.
Moving on to part B, when 2 mL of 1 M NaOH is added to the solution in part A, a neutralization reaction occurs. NaOH is a strong base and reacts with TRIS-H+ to form the salt TRIS-Na+ and water. This reaction consumes the TRIS-H+ and increases the concentration of TRIS-Na+ in the solution.
Since the TRIS-H+ is being consumed, the pH of the solution will increase. However, to determine the exact pH, we need to calculate the new concentration of TRIS and TRIS-Na+ in the solution.
To find the moles of TRIS in the solution after the addition of NaOH, we'll use the initial moles of TRIS minus the moles of TRIS that reacted with NaOH. NaOH has a 1:1 stoichiometry with TRIS-H+ (acid), so 2 mmol (0.002 moles) of TRIS will react with 2 mmol of NaOH.
Therefore, the remaining moles of TRIS in the solution = initial moles of TRIS - moles of TRIS reacted with NaOH = 0.005 moles - 0.002 moles = 0.003 moles.
Now, we can calculate the concentration of TRIS in the updated solution by dividing the moles of TRIS by the volume of the solution (52 mL or 0.052 L):
Molarity = 0.003 moles / 0.052 L ≈ 0.0576 M
Since the solution now contains TRIS-Na+ as well, we need to account for its contribution to the pH calculation. However, TRIS-Na+ is a neutral salt and does not affect the pH significantly.
Therefore, the pH of the solution after the addition of 2 mL of 1 M NaOH will be very close to the pH of the initial solution, which is 8.3.