The document discusses calculating the concentration of hydroxide ions after mixing solutions of aluminum nitrate and sodium hydroxide. It explains that 100 mL of 0.1 M aluminum nitrate is mixed with 100 mL of 0.4 M sodium hydroxide, resulting in 10 mmol of aluminum ions reacting with 30 mmol of hydroxide ions to form aluminum hydroxide precipitate. This leaves 10 mmol of hydroxide ions in the total 200 mL solution volume, giving a concentration of hydroxide ions of 0.05 M.

1. I want to calculate concentration of hydroxide ion, but why is it 10 milimole / 200ml?
shouldn't it be 20 or 40 milimole?
12. Suppose that 100.0 mL of 0.1000 M AI(NOs)3 is mixed with 100.0 mL of 0.4000 M NaOH.
Solution
Here Al(OH)3 precipitate is being formed
mol of Al3+ added = 0.100 M * 100 mL
= 10 mmol
Al3+  +  3 OH- —> Al (OH)3
So,
10 mmol of Al3+ will react with 30 mmol of OH- to form precipiatye
So,
remaining mol of OH- = 40 mmol - 30 mmol = 10 mmol
total volume = 200 mL
This is the reason,
concentration of OH- = 10 mmol / 200 mL =
= 0.050 M
Answer: 0.050 M