Chemistry Labs

Problem 3

Silver chloride is a milk-white solid (quotes from a lesson by L. J. Gay-Lussac). Data at 298 K: pKs1(AgCl)=9.7pK_{s1}(\ce{AgCl}) = 9.7; pKs2(AgX2CrOX4)=12pK_{s2}(\ce{Ag2CrO4}) = 12; formation constant of [Ag(NHX3)Xn]+[\ce{Ag(NH3)_n}]^+: βn=107.2\beta_n = 10^{7.2}; E∘(AgX+/Ag)=0.80E^\circ(\ce{Ag+/Ag}) = 0.80 V. (a) Calculate the solubility ss of AgCl(s)\ce{AgCl(s)} in water. (b) When ammonia is added to silver chloride a complex of stoichiometry nn is formed; write the equilibrium and its constant KK, then determine nn knowing that 0.10.1 mol of AgCl\ce{AgCl} in 1 dm3^3 of water just dissolves when [NHX3]=1.78[\ce{NH3}] = 1.78 mol dm−3^{-3}. (c) The Mohr method titrates ClX−\ce{Cl-} by AgX+\ce{Ag+} in the presence of KX2CrOX4\ce{K2CrO4}: three drops (≈0.5\approx 0.5 cm3^3) of KX2CrOX4\ce{K2CrO4} at c=7.76×10−3c = 7.76\times 10^{-3} mol dm−3^{-3} are added to V0=20.00V_0 = 20.00 cm3^3 of a NaCl solution titrated by AgNOX3\ce{AgNO3} at c=0.050c = 0.050 mol dm−3^{-3}; a red precipitate appears at VAg=4.30V_{\ce{Ag}} = 4.30 cm3^3. Calculate c(ClX−)c(\ce{Cl-}) and the residual [ClX−]res[\ce{Cl-}]_{\text{res}} when AgX2CrOX4\ce{Ag2CrO4} starts to precipitate.
Step 1 of 5: Solubility of AgCl
s=Ks1=10−9.7=1.41×10−5 mol dm−3s = \sqrt{K_{s1}} = \sqrt{10^{-9.7}} = 1.41\times 10^{-5}\ \text{mol dm}^{-3}
Analysis

For AgCl(s)⇌AgX++ClX−\ce{AgCl(s) <=> Ag+ + Cl-}, [AgX+]=[ClX−]=s[\ce{Ag+}] = [\ce{Cl-}] = s so Ks1=s2K_{s1} = s^2 and s=1.41×10−5s = 1.41\times 10^{-5} mol dm−3^{-3}.