Respuesta :
Answer:
The pH of the buffer system is 9.57
Explanation:
Step 1: Data given
Molarity of NH4Cl = 0.045 M
Molarity of NH3 = 0.095 M
Ka of ammonium is 5.6*10^-10
pKa = -log (5.6 * 10^-10) = 9.25
Step 2: The balanced equation
NH4+(aq) ⇆ H+(aq) + NH3(aq)
Step 3: Calculate the pH of the buffer
pH = pKa + log[NH3]/[NH4+]
pH = 9.25 + log(0.095/0.045)
pH = 9.57
The pH of the buffer system is 9.57
Answer:
9.57
Explanation:
pH = pKa + log([NH3]/[NH4CL])
pKa = -logKa = -log(5.6×10^-10) = 9.25
[NH3] = 0.095M
[NH4CL] = 0.045M
pH = 9.25 + log(0.095/0.045) = 9.25 + log2.11 = 9.25 + 0.32 = 9.57