the groundwater well used for water supply of City of Nowhere, USA contain Answe

Question

the groundwater well used for water supply of City of Nowhere, USA contain

Answer all questions. Nowhere, USA contains 5.3 mgyl of 1. The groundwater well used for water supply of City of aluminum (Al' The city wants to lower the aluminum levels to less than 0.5 mg/l. How much Aluminum will stay in solution if the pH is reduced to 4.5. (i) on the other hand id the city wants to meet the goal of 0.5mg/l for aluminum, what pH is required. The following (0) equation governs precipitation. Al 3 3 OH, Ks

Explanation / Answer

SOLUTION:

Al(OH)3 <--------> Al3+ + 3OH-

Ks = [OH-]3[Al3+ ]

let s is the Molar solubility

10-32.9 = [s]3[s ] = [s]4

(i) When pH = 4.5

pOH = 14 - 4.5 = 9.5

Molarity of OH- = antilog(-pOH) = 3.16 X 10-10 M

10-32.9 = [OH-]3[Al3+ ]

10-32.9 = [3.16 X 10-10 ]3[Al3+ ]

[Al3+ ] = 10-32.9 / [3.16 X 10-10 ]3 = 10-32.9 / 31.55X 10-30 = 3.96 X 10-5M

Moles of Al3+ = Molarity X Volume = 3.96 X 10-5 Moles/L X 1L = 3.96 X 10-5 Moles

Take the volume as 1L

Mass of Al3+ in 1L of water = Molar mass X Number of moles

= 27g X 3.96 X 10-5 Moles = 1.06 X 10-3g = 1.06mg

(ii) If the city want 0.5mg/L

Convert the 0.5mg/L to malarity

Molarity = number of moles / Volume

Number of moles = mass / molar mass = 0.5 X 10-3 / 27; Volume = 1L

Molarity = 0.5 X 10-3 / 27 X 1L = 1.85 X 10-5M

Insert this value in Ks equation

Ks = [OH-]3[Al3+ ]

10-32.9 = [OH-]3[1.85 X 10-5]

[OH-]3 = 10-32.9 / 1.85 X 10-5 = 6.76 X 10-29

[OH-] = (6.76 X 10-29)1/3 = 4.07 X 10-10M

pOH = -log(4.07 X 10-10) = 9.39

pH = 14 - 9.39 = 4.61

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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