Answer:
pH = 4.93
Explanation:
Number of moles of weak base (NH3) = 110 × 0.5568 = 61.248
Number of moles of acid (HCl) = 0.4412 × 138.32 = 61.248
At equivalence point we have M1V1 = M2V2
00.4412 × V1 = 110 × 0.5568
V1 = 138.82 ml
We have pKw = pH +pOH
pOH = 1/2(pKb + pKw + Log Cs) = 1/2 (4.75 + 14 + log (61.248(138.82+110))
pOH = 1/2(4.75+14+log 0.25)
pOH = 9.075
So pH = pkw-pOH = 14 - 9.075 = 4.925 = 4.93. Therefore, the pH at equivalence is 4.93.
Answer:
pH = 4.93
Explanation:
Step 1: Data given
Volume of a 0.5568 M ammonia NH3 solution = 110.0 mL = 0.110 L
Molarity of HCl = 0.4412 M
pKb ammonia = 4.75
Step 2: The balanced equation
NH3 + HCl ⇒ NH4+ + Cl-
Step 3: Calculate moles NH3
Moles NH3 = molarity NH3 * volume
Moles NH3 = 0.5568 M * 0.110 L
Moles NH3 = 0.06125 moles
Step 3: Calculate moles HCl needed
For 0.06125 moles NH3 we need 0.06125 moles HCl
Step 4: Calculate volume HClFor 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4+
Volume HCl = moles HCl / molarity
Volume HCl = 0.06125 moles / 0.4412 M
Volume HCl = 0.139 L = 139 mL
Step 5: Calculate total volume
Total volume = 110.0 mL + 139 mL
Total volume = 249 mL
Step 6: Calculate moles NH4+
For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4+
For 0.06125 moles NH3 we'll have 0.06125 moles NH4+
Step 7: Calculate [NH4+]
[NH4+] = moles NH4+ / volume
[NH4+] = 0.06125 moles / 0.249 L
[NH4+]= 0.246 M
Step 8: The equation
NH4+ + H2O → NH3 + H3O+
Step 9: The initial concentration
[NH4+] = 0.246 M
[NH3]= 0M
[H30+] = 0M
Step 10: Calculate the concentration at the equilibrium
[NH4+] = 0.246 - X M
[NH3]= X M
[H30+] = X M
Step 11: Calculate Ka
pKa = 14 - pKb
pKa = 14 - 4.75 = 9.25
Ka = 10^-9.25 = 5.6 *10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka = X² / (0.246 - X)
5.6 *10^-10 = X² / (0.246 - X)
5.6 *10^-10 = X² / (0.246)
X = 1.17*10^-5 M = [ H3O+]
Step 12: Calculate pH
pH = - log [1.17*10^-5]
pH = 4.93
