Consider the titration of 100.0 mL 0.1 M KOH with 1.0 M HBr. Find the pH at the following volumes of acid (Va) added: Va= 0, 1, 10 and 10.1 mL.
First determine the volume for the equivalence point.
For zero mL you have a solution of KOH only. Since that is a strong base and is 100% ionized, solve for OH and convert to pH.
At the equivalence point, the pH is determined by the hydrolysis of the salt.
At all points before the equivalence point, pH is due to excess OH^-.
At all point after the equivalence oint the pH is determined by the excess HBr.
For before and after the equivalence point points, remember to take into account the dilution of the KOH or HBr.
To find the pH at the given volumes of acid added during the titration, we need to determine the moles of acid and base present at each step and use the amount of base and acid to calculate the resulting concentration of hydroxide ions (OH-) and hydronium ions (H3O+). The pH can then be calculated from the concentration of hydronium ions using the formula -log[H3O+].
First, let's determine the moles of KOH present initially:
Moles of KOH = volume of KOH (L) × concentration of KOH (mol/L)
= 0.100 L × 0.1 mol/L
= 0.010 mol
At Va = 0 mL (initially), the moles of HBr added is 0. Therefore, the moles of KOH remaining is still 0.010 mol. Since the KOH is a strong base, it will completely dissociate in water, producing equal moles of hydroxide ions (OH-).
Concentration of OH- = moles of OH- / total volume of solution (L)
= 0.010 mol / 0.100 L
= 0.100 M
Now, to calculate the pOH at V_a = 0 mL:
pOH = -log[OH-]
= -log[0.100]
= 1.000
pOH + pH = 14 (at 25°C). Therefore, the pH at V_a = 0 mL is:
pH = 14 - pOH = 14 - 1.000 = 13.000
Next, let's consider V_a = 1 mL. At this point, 1 mL of 1 M HBr is added. To calculate the moles of HBr added:
Moles of HBr = volume of HBr (L) × concentration of HBr (mol/L)
= 0.001 L × 1.0 mol/L
= 0.001 mol
Since HBr is a strong acid, it will completely dissociate in water, producing equal moles of hydronium ions (H3O+). Therefore, the moles of H3O+ at V_a = 1 mL is also 0.001 mol.
Now, let's calculate the moles of KOH remaining:
Moles of KOH remaining = initial moles of KOH - moles of H3O+
= 0.010 mol - 0.001 mol
= 0.009 mol
Now, calculate the volume of the solution after V_a = 1 mL:
Total volume = initial volume (0.100 L) + volume of acid added (0.001 L)
= 0.101 L
Concentration of OH- = moles of OH- / total volume of solution (L)
= 0.009 mol / 0.101 L
= 0.089 M
Similarly, calculate pOH using the concentration of OH-:
pOH = -log[OH-]
= -log[0.089]
= 1.050
pH = 14 - pOH = 14 - 1.050 = 12.950
Repeat this process to find the pH at V_a = 10 mL and V_a = 10.1 mL by finding the moles of acid added, moles of base remaining, and concentration of OH- at each step. Then calculate the pOH and pH using the same formulas.
Remember that the concentration of OH- will decrease as more acid is added, resulting in a decrease in pH.