How many grams of dry NH4Cl need to be added to 1.60L of a 0.700\\it M solution

Question

How many grams of dry NH4Cl need to be added to 1.60L of a 0.700it M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 9.00? K_b for ammonia is 1.8times 10^{-5}.

Explanation / Answer

NH4Cl ----------> NH4(+) + Cl(-) NH3 + H2O NH4(+) + OH(-) The Kb is small so the it is possible to say that all the NH4(+) is released from NH4Cl and the concentration of NH3 will not change . C of NH3 = 0.6 M pOH = 14 - 8.77 C of OH(-) = 10 ^ ( - pOH ) 1.8 * 10^-5 = [ C(OH) * C(NH4) ] / C(NH3) grams of NH4Cl = [ 1.6 liters ] * [ C(NH4) ] * [ molecular waigh of NH4Cl ]

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.