Question
Download Solution PDFWhat is the molar solubility of Al(OH)3 in 0.2 M NaOH Solution? Given that, solubility product Al(OH)3 = 2.4 × 10-24:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Let the solubility of
Al(OH)3 ⇌ Al3+ + 3OH-
s 3s
NaoH ⇌ Na++ OH-
0.2 M 0.2M
Calculation:
[(Al3+)] = s and [OH-] = 3s + 0.2 ≈ 0.2
ksp = 2.4 × 10-24 = [Al3+] [OH-]
2.4 × 10-24 = s(0.2)3
\(s = \frac{{2.4 \times {{10}^{ - 24}}{\rm{\;}}}}{{8 \times {{10}^3}{\rm{\;}}}} = 3 \times {10^{ - 22}}\;{\rm{mol}}/L\)Last updated on Jul 11, 2025
-> JEE Main 2026 application will start probably from second week of October 2025 till November 2025.
->Check JEE Main syllabus for 2026 examination.
-> JEE Main is a national-level engineering entrance examination conducted for 10+2 students seeking courses B.Tech, B.E, and B. Arch/B. Planning courses.
-> JEE Mains marks are used to get into IITs, NITs, CFTIs, and other engineering institutions.
-> All the candidates can check the JEE Main Previous Year Question Papers, to score well in the JEE Main Exam 2025.