a.
pH = 0.602
b.
pH = 1.5
c.
pH = 7
d.
pH = 12.1
Explanation:
a ) To calculate the pH, use the following equation:
pH = -log [H+]
Hbr is a strong acid, so the [H+] concentration can be calculated as follow:
[HBr] = 0.250 M
As acid Hbr:
Hbr = H+ + Br-
As strong acid HBr dissociates at all, so
[HBr] = [H+] = 0.250 M
So the pH:
pH = -log [0.250 M] = 0.602
b) Calculate the pH after adding 250 mL of 0.125M NaOH
In this point, the reactions starts:
HBr + NaOH = H2O + NaBr
- First, we gonna find the mol of each reactant:
HBr:
mol = [M] × L
mol = 0.250 M × 0.250 L
mol = 0.0625 mol HBr
NaOH:
mol = [M] × L
mol = 0.125 M × 0.250L
mol = 0.03125 mol NaOH
In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:
0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr
These are the moles free in the solution, and we going to use them to calculate the pH:
pH = -log [ H]
pH = - log [ 0.03125 ] = 1.5
c) Calculate the pH after adding 500 mL of 0.125M NaOH
NaOH:
mol = [M] × L
mol = 0.125 M × 0.500L
mol = 0.0625 mol NaOH
In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:
0.0625 mol HBr – 0.0625 mol HBr = 0 HBr
pH = 7
d) Calculate the pH after adding 600 mL of 0.125M NaOH.
NaOH:
mol = [M] × L
mol = 0.125 M × 0.600L
mol = 0.075 mol NaOH
In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:
0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH
These are the moles free in the solution, and we going to use them to calculate the pH:
pOH = -log [ OH]
pOH = - log [ 0.0125 ] = 1.90
pH + pOH = 14
pH = 14- pOH = 14 – 1.90 = 12.1
