What is the pH that would result from the addition of 0.002 moles of HCl to 100 mL of a 0.1 M boric acid/borate, pH= 9.2 buffer solution?
Use the Henderson-Hasslebalch equation.
Sucks to be you!
To determine the resulting pH after adding HCl to a buffer solution, you need to consider the reaction between the HCl and the components of the buffer system. In this case, the buffer system consists of boric acid (H3BO3) and its conjugate base, borate (B(OH)4-).
Step 1: Write the balanced chemical equation for the reaction between HCl and borate (B(OH)4-):
HCl + B(OH)4- → H2BO3- + Cl-
Step 2: Determine the initial moles of the buffer components in the 100 mL solution:
Initial moles of H3BO3 = (0.1 M) × (100 mL) = 0.01 moles
Initial moles of B(OH)4- = 0 moles
Step 3: Determine the limiting reagent. Since the moles of HCl added are given as 0.002 moles, HCl will be the limiting reagent.
Step 4: Determine the moles of H3BO3 and B(OH)4- consumed based on the stoichiometry of the reaction. From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of B(OH)4-. Therefore, 0.002 moles of HCl will react with 0.002 moles of B(OH)4-.
Step 5: Calculate the moles of H3BO3 and B(OH)4- remaining in the solution:
Moles of H3BO3 remaining = Initial moles of H3BO3 - Moles of H3BO3 consumed = 0.01 moles - 0.002 moles = 0.008 moles
Moles of B(OH)4- remaining = Initial moles of B(OH)4- - Moles of B(OH)4- consumed = 0 moles - 0.002 moles = -0.002 moles
Step 6: Convert the moles of H3BO3 remaining to concentration (M) in the final volume of the solution (100 mL + volume of HCl added):
Concentration of H3BO3 remaining = Moles of H3BO3 remaining / Total volume (L)
= 0.008 moles / 0.1 L
= 0.08 M
Step 7: Use the Henderson-Hasselbalch equation to calculate the resulting pH:
pH = pKa + log ([A-] / [HA])
In this case, H3BO3 is the conjugate acid (HA) and B(OH)4- is the conjugate base (A-).
pKa is the dissociation constant of H3BO3.
The pKa of H3BO3 is 9.24.
pH = 9.24 + log ([B(OH)4-] / [H3BO3])
Since [B(OH)4-] = 0.002 moles (from the moles consumed in Step 4) and [H3BO3] = 0.008 moles (from Step 6):
pH = 9.24 + log (0.002 / 0.008)
= 9.24 + log (0.25)
≈ 9.09
So, the resulting pH after adding 0.002 moles of HCl to 100 mL of the buffer solution is approximately 9.09.