solution of 0.1molar NH4OH and 0.1 NH4CL ph=9.25 calculate kb for NH4OH?
i also need the answer
4.75
To calculate the value of K_b for NH4OH, you need to use the pH and the concentrations of NH4OH and NH4Cl. Here are the steps to determine K_b:
Step 1: Write the balanced chemical equation for the dissociation of NH4OH:
NH4OH (aq) ⇌ NH4+ (aq) + OH- (aq)
Step 2: Write the expression for the equilibrium constant (K_b) for the dissociation of NH4OH:
K_b = [NH4+][OH-] / [NH4OH]
Step 3: Determine the initial concentrations of NH4OH and NH4Cl:
Given that NH4OH is 0.1 M (0.1 mol/L) and NH4Cl is 0.1 M.
Step 4: Calculate the concentration of OH- (hydroxide ions) using the pH:
pOH = 14 - pH
pOH = 14 - 9.25 (since pH is given as 9.25)
Step 5: Calculate the concentration of OH- using the pOH:
OH- concentration = 10^(-pOH)
OH- concentration = 10^(-4.75) (using the calculated pOH value)
Step 6: Calculate the concentrations of NH4+:
Since the NH4OH and NH4Cl are both strong electrolytes, they will dissociate completely in water:
NH4OH: 0.1 M × 1 = 0.1 M
NH4Cl: 0.1 M × 1 = 0.1 M
Step 7: Substitute the values into the K_b expression:
K_b = [NH4+][OH-] / [NH4OH]
K_b = (0.1 M) × (OH- concentration) / (0.1 M)
Step 8: Simplify the expression:
K_b = OH- concentration
Step 9: Calculate the value of K_b:
K_b = OH- concentration = 10^(-4.75)
After performing the calculations, you will calculate the value of K_b for NH4OH as 5.623 × 10^(-5).
pH = pKa + log (base)/(acid)
Substitue and solve for pKa, then convert pKa to pKb, then to Kb
pKa + pKb = pKw = 14.
and pKb = -logKb