HMMT 十一月 2017 · THM 赛 · 第 3 题

HMMT November 2017 — THM Round — Problem 3

E milia wishes to create a basic solution with 7% hydroxide (OH) ions. She has three solutions of different bases available: 10% rubidium hydroxide (Rb(OH)), 8% cesium hydroxide (Cs(OH)), and 5% francium hydroxide (Fr(OH)). (The Rb(OH) solution has both 10% Rb ions and 10% OH ions, and similar for the other solutions.) Since francium is highly radioactive, its concentration in the final solution should not exceed 2%. What is the highest possible concentration of rubidium in her solution?

解析

E milia wishes to create a basic solution with 7% hydroxide (OH) ions. She has three solutions of different bases available: 10% rubidium hydroxide (Rb(OH)), 8% cesium hydroxide (Cs(OH)), and 5% francium hydroxide (Fr(OH)). (The Rb(OH) solution has both 10% Rb ions and 10% OH ions, and similar for the other solutions.) Since francium is highly radioactive, its concentration in the final solution should not exceed 2%. What is the highest possible concentration of rubidium in her solution? Proposed by: Yuan Yao Answer: 1% Suppose that Emilia uses R liters of Rb(OH), C liters of Cs(OH), and F liters of Fr(OH), then we have 10% · R + 8% · C + 5% · F 5% · F = 7% and ≤ 2% . R + C + F R + C + F The equations simplify to 3 R + C = 2 F and 3 F ≤ 2 R + 2 C , which gives 9 R + 3 C ≤ 2 R + 2 C ⇒ 5 R ≤ C. 2 Therefore the concentration of rubidium is maximized when 5 R = C , so F = 4 R , and the concentration of rubidium is 10% · R = 1% . R + C + F