When 23.3 mL of 0.14 M HCl are added to 50.0 mL of a 0.16 M solution of a weak monoprotic base, the pH of the solution is 10.50. What is the Kb of the weak base?
do u find moles of both and then add the moles for HB+ and then just use the B- moles for the B on the henderson hasselbach equation
and then solve for Ka and then solve for Kb?
i don't get the right answer
please help
Find moles of both, SUBTRACT mols acid from moles base (because base is larger), then plug into the HH equation to calculate pKa, change to pKb, then to Kb.
What do you plug into the HH equation?
To find the Kb of the weak base, you need to follow a step-by-step approach. Let's break it down:
Step 1: Calculate the moles of HCl added.
To determine the moles, you need to multiply the volume (23.3 mL) by the molarity (0.14 M):
Moles of HCl = 23.3 mL * 0.14 mol/L = 3.262 mol
Step 2: Determine the initial moles of the weak base (B-).
To find the initial moles, multiply the volume (50.0 mL) by the molarity (0.16 M):
Moles of weak base = 50.0 mL * 0.16 mol/L = 8.000 mol
Step 3: Calculate the moles of HCl reacting with the weak base.
Since HCl and weak base react in a 1:1 ratio, the moles of weak base reacting with HCl are also 3.262 mol.
Step 4: Determine the remaining moles of the weak base.
Subtract the moles of HCl reacting with the weak base from the initial moles of the weak base:
Remaining moles of weak base = Initial moles of weak base - Moles of weak base reacting with HCl
Remaining moles of weak base = 8.000 mol - 3.262 mol = 4.738 mol
Step 5: Calculate the concentration of the weak base [B-].
Divide the remaining moles of weak base by the total volume in liters:
[B-] = Moles of weak base / Total volume (in liters)
[B-] = 4.738 mol / (50.0 mL + 23.3 mL) = 0.07656 M
Step 6: Calculate the pOH.
Use the pH of the solution to find the pOH. Since pH + pOH = 14, subtract the pH from 14:
pOH = 14 - 10.50 = 3.50
Step 7: Calculate the concentration of [OH-].
Convert the pOH to [OH-] using the equation [OH-] = 10^(-pOH):
[OH-] = 10^(-pOH) = 10^(-3.50) = 3.16 x 10^(-4)
Step 8: Calculate the concentration of [H+] using the equation [H+] = 1.0 x 10^(-14) / [OH-].
[H+] = 1.0 x 10^(-14) / [OH-] = 1.0 x 10^(-14) / 3.16 x 10^(-4) = 3.16 x 10^(-11)
Step 9: Calculate the concentration of weak acid [HB+] since weak base will only partially convert to HB+ and [OH-].
[H+] = [B-]
[HB+] + [OH-] = X (Since [OH-] converts to [HB+])
Using the Henderson-Hasselbalch equation:
pOH = pKa + log([B-]/[HB+])
3.50 = pKa + log([B-]/[HB+])
log([B-]/[HB+]) = 3.50 - pKa
Step 10: Calculate the Kb.
Kw = Ka * Kb (Water dissociation constant)
1.0 x 10^(-14) = Ka * Kb
From Step 9, we know that:
log([B-]/[HB+]) = 3.50 - pKa
Let's assume pKa is a given value.
Using the information above, let's analyze if step 9 and step 10 are correct. Could you please reconfirm the concentration of [OH-]?