Question

which solution has the lowest concentration of ions present?
o 1.0 m agcl
o 1.0 m naf
o 1.0 m k₂so₄
o 1.0 m koh

Explanation:

Step1: Analyze the dissociation of each compound

• For \(KOH\): \(KOH = K^{+}+OH^{-}\), so the total ion concentration \(c = 1.0 + 1.0=2.0M\)
• For \(K_{2}SO_{4}\): \(K_{2}SO_{4}=2K^{+}+SO_{4}^{2 -}\), so the total ion concentration \(c = 2\times1.0+1.0 = 3.0M\)
• For \(NaF\): \(NaF = Na^{+}+F^{-}\), so the total ion concentration \(c = 1.0+1.0 = 2.0M\)
• For \(AgCl\): \(AgCl\) is a sparingly - soluble salt. Its solubility product \(K_{sp}=1.8\times10^{-10}\). Let \(s\) be the solubility of \(AgCl\) in water. From \(K_{sp}=[Ag^{+}][Cl^{-}]=s\times s\), we get \(s=\sqrt{K_{sp}}=\sqrt{1.8\times 10^{-10}}\approx1.34\times10^{-5}M\). The total ion concentration \(c = s + s\approx2.68\times10^{-5}M\)

Answer:

\(1.0M\ AgCl\)