Given a 300 ml solution containing 0.2M H2L+ calculate the pH after the addition of 130.00 ml of 0.6M KOH.
-----pH = 9.377
What total volume of KOH must be added to reach pH of 10.00
---0.386 moles
However, I am getting an amount of ml that is way too high, according to the graph it should be between 150-200ml...
Please help!
What are you using for pK2 for H2L? It might help if I knew pK1 also.
To calculate the total volume of KOH needed to reach a pH of 10.00, we need to use the concept of acid-base titration. In this case, we are adding a strong base (KOH) to a weak acid (H2L+), and the pH will increase as more KOH is added.
First, let's calculate the initial moles of H2L+ in the solution:
moles of H2L+ = concentration of H2L+ * volume of solution
= 0.2 M * 300 mL
= 0.06 moles of H2L+
Next, we need to determine the moles of KOH added to the solution.
moles of KOH = concentration of KOH * volume of KOH
= 0.6 M * 130 mL
= 0.078 moles of KOH
Since KOH is a strong base, it will fully react with H2L+ according to the balanced chemical equation:
H2L+ + OH- -> HL + H2O
From the equation, we see that every mole of H2L+ reacts with one mole of OH-. Therefore, the moles of KOH equal the moles of OH-.
To calculate the new concentration of OH-:
moles of OH- = 0.078 moles
Next, we need to calculate the new volume of the solution after the addition of KOH:
new volume = initial volume + volume of KOH
= 300 mL + 130 mL
= 430 mL
Given that pOH = -log[OH-], we can calculate the pOH using the following equation:
pOH = -log (concentration of OH-)
To convert pOH to pH, we can use the equation:
pH = 14 - pOH
Now, we can calculate the new pH after the addition of the KOH:
pH = 14 - (-log (0.078))
= 14 - 1.107
= 12.893
To reach a pH of 10.00, we need to neutralize the excess OH-. Therefore, we can use the following equation:
moles of OH- = moles of H2L+
Since we know the moles of H2L+ is 0.06, we can set up the equation:
0.06 moles = concentration of KOH * total volume of KOH
Rearranging the equation to solve for the total volume of KOH:
total volume of KOH = 0.06 moles / concentration of KOH
= 0.06 moles / 0.6 M
= 0.1 L = 100 mL
Based on the calculations, the total volume of KOH needed to reach a pH of 10.00 is 100 mL, which is significantly lower than the initial estimation of 150-200 mL. Please double-check your calculations and ensure that you're using the correct units and formulas.