What is the pH of a buffer solution prepared by mixing 100.0 mL of 0.200 mol L^-

Question

What is the pH of a buffer solution prepared by mixing 100.0 mL of 0.200 mol L^-1 NH_3 and 85.7 mL of 0.250 mol L^-1 NH_4C1? K_b(NH_3) = 1.76 times 10^-5 A) 4.725 B) 9.275 C) 4.245 D) 5.658 E) 9.216 28. What is the pH of a 30.0 mL buffer solution composed of 0.30 mol L^-1 NH_3 and 0.30 mol L^-1 NH_4C1 after 0.0015 mol of NaOH(s) are added to it at 25 degreeC? Assume that the volume of the solution does not change by the addition of the solid. K_b(NH_3) = 1.76 times 10^-5. A) 4.901 B) 9.099 C) 8.654 D) 9.392 E) 4.754

Explanation / Answer

27) Ans= E

[NH4Cl] = 85.7 mL x 0.25 M = 21.425 mmol

[ NH3] = 100 mL x 0.2 M = 20 mmol

Kb = 1.76 x 10-5

pOH = - logKb + log [NH4Cl]/[NH3]

= -log(1.76 x 10-5) + log (21.425 mmol/ 20 mmol)

= 4.78

pH = 14- pOH = 14 -4.78 = 9.216

Therefore,

pH = 9.216

28) NH4Cl + NaOH ----------> NH3 + H2O + NaCl

Given that

[NH4Cl] = 30 mL x 0.3 M = 9 mmol

[ NH3] = 30 mL x 0.3 M = 9 mmol

[NaOH] = 0.0015 mol = 0.0000015 mmol

Hence,

pOH = - logKb + log [NH4Cl]- [NaOH]/[NH3] + [NaOH]

= -log(1.76 x 10-5) + log ( 9 mmol -0.0000015 mmol / 9 mmol + 0.0000015 mmol)

= 4.75

pH = 14 -pOH = 14-4.75 = 8.25

pH = 8.25

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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