Question

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15. If you make up a solution of 100mL of 0.15 Tris-Base, what will be the pH? (2 pts oll- oool x ²0l = zvolomillin ni+H 16. If you add 4mL of 1M of HCl to the solution in question 15, what will be the pH? (4 pts) How 1000 lomlox Jool. = 10H Hloto Jmool 2-01 x 24 e gone oss ox jogoo (HO) NO Morso do Smod. 10M 21 0 0 0 1 x 27 x S oHo do som 356 02 (10.0) (210.0)HO 2938 SOM200.0 = 220009 ENOJ HCO

Answer 1

15) You have the reaction:

Tris + H2O = Tris + + OH-

The expression of Kb:

Kb = [Tris +] * [OH-] / [Tris]

8.7x10 ^ -9 = X ^ 2 / 0.15 - X

It is assumed that - X is negligible and clears:

X = [OH-] = √8.7x10 ^ -9 * 0.15 = 3.6x10 ^ -5 M

The pOH and the pH are calculated:

pOH = - log 3.6x10 ^ -5 = 4.4

pH = 14 - 4.4 = 9.6

16) The added moles of HCl are calculated:

n HCl = M * V = 1 M * 0.004 L = 0.004 mol

The moles of Tris base are calculated:

n Tris = 0.15 M * 0.1 L = 0.015 mol

The HCl reacts with Tris (decreasing it) and forms Tris + (increasing it), the pH is calculated by the equation of henderson hasselbach:

pH = pKa + log (n Tris / n Tris +) = 8.06 + log (0.015 - 0.004 / 0.004) = 8.5

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