Final answer:
The pH of the buffer system made of 0.15 M NH3 and 0.35 M NH4Cl is calculated using the Henderson-Hasselbalch equation, considering NH4+ as the conjugate acid of NH3 with a pKa of 9.25.
Step-by-step explanation:
To calculate the pH of the buffer system made of 0.15 M NH3 and 0.35 M NH4Cl, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([Base]/[Acid])
For the weak base ammonia (NH3), the value of Kb is 1.8×10-5, implying that the Ka for the dissociation of its conjugate acid, NH4+, is Kw/Kb = 10-14/1.8×10-5 = 5.6×10-10. Thus, the pKa for NH4+ is 9.25. Now we can apply the Henderson-Hasselbalch equation:
pH = 9.25 + log(0.15/0.35)
pH = 9.25 + log(0.4286)
pH = 9.25 - 0.3676
pH = 8.8824
The calculated pH of the buffer solution is approximately 8.88. The buffer is therefore slightly basic since its pH is above 7.
The pH of a buffer system can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, NH₃ acts as the base (A-) and NH₄Cl acts as the acid (HA). The pKa value for the NH₄+/NH₃ system is 9.25. Substituting the given concentrations into the equation, we get:
pH = 9.25 + log(0.15/0.35) = 8.91
Therefore, the pH of the buffer system made up of 0.15 M NH₃ and 0.35 M NH₄Cl is 8.91.