Recently, one of Secondary 4 students asked me a question on Salts Solubility that is related to Mole Calculations, and i would like to share with you on the thought process to solve such a question.
Question:
Excess amount of water was added to a mixture consisting of sodium chloride, sodium sulfate and silver chloride salts. The solution that is formed is known to contain 0.4 mol of chloride ions and 0.6 mol of sulfate ions. What is the number of moles of sodium ions present in the solution?
A) 0.4 mol
B) 1.0 mol
C) 1.4 mol
D) 1.6 mol
Now, stop scrolling and start trying out the question first, before you scroll down to look at the suggested solution!
Suggested Solution:
This question is testing students on two fundamental concepts:
- Solubility of Common Salts
- Molar Ratio of Ions upon Dissociation of Ionic Compounds
Based on the Table of Solubility of Common Salts, sodium chloride and sodium sulfate are soluble in water and will dissociate into its ions.
Silver chloride is an insoluble salt and will not dissociate at all i.e. no ions produced when it is placed inside water.
NaCl(aq) → Na+(aq) + Cl–(aq)
1 mol of NaCl will dissociate and produce 1 mol of Na+ ions & 1 mol of Cl– ions.
Since 0.4 mol of Cl– ions is produced, 0.4 mol of Na+ ions will also be produced from dissociation of NaCl.
Na2SO4(aq) → 2Na+(aq) + SO42-(aq)
1 mol of Na2SO4 will dissociate and produce 2 mol of Na+ ions and 1 mol of SO42- ions.
Since 0.6 mol of SO42- ions is produced, 0.6 x 2 = 1.2 mol of Na+ ions will be produced from dissociation of Na2SO4.
As such, total no. of moles of Na+ ions present in solution = 0.4 + 1.2 = 1.6 mol
Hence, answer is D.
Hope you learned something useful from this discussion.
Keep learning and stay tune for the next post.
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