Chemistry Labs

Problem 4

Gold Capital of Asia. Chiufen, northeast Taiwan, was one of the largest gold mines in Asia. PART A. KCN extracts gold from ore: 4 Au(s)+8 CNX−(aq)+OX2(g)+2 HX2O(l)→4 [Au(CN)X2]X−(aq)+4 OHX−(aq)\ce{4Au(s) + 8CN-(aq) + O2(g) + 2H2O(l) -> 4[Au(CN)2]-(aq) + 4OH-(aq)}. (4.A-1) State the geometry of [Au(CN)X2]X−\ce{[Au(CN)2]-}. (4.A-2) How many grams of KCN are needed to extract 20 g of gold? (M(Au)=197M(\ce{Au}) = 197, M(KCN)=65.12M(\ce{KCN}) = 65.12 g mol−1^{-1}.) Aqua regia (3:1 conc. HCl:HNOX3\ce{HNO3}) dissolves gold by redox: Au(s)+3 NOX3X−(aq)+6 HX+(aq)+4 ClX−(aq)→[AuClX4]X−(aq)+3 NOX2(g)+3 HX2O(l)\ce{Au(s) + 3NO3-(aq) + 6H+(aq) + 4Cl-(aq) -> [AuCl4]-(aq) + 3NO2(g) + 3H2O(l)}. (4.A-3) Write the half-reactions and the balanced equation. (4.A-4) Identify the oxidizing and reducing agents. (4.A-5) Given AuX3++3 eX−→Au(s)\ce{Au^{3+} + 3e- -> Au(s)} (E∘=+1.50E^\circ = +1.50 V) and [AuClX4]X−+3 eX−→Au(s)+4 ClX−\ce{[AuCl4]- + 3e- -> Au(s) + 4Cl-} (E∘=+1.00E^\circ = +1.00 V), calculate the formation constant Kf=[[AuClX4]X−][AuX3+][ClX−]4K_f = \frac{[\ce{[AuCl4]-}]}{[\ce{Au^{3+}}][\ce{Cl-}]^4} at 25 ∘C25\ ^\circ\mathrm{C} (RTln⁡10/F=0.059RT\ln10/F = 0.059 V). PART B. Gold nanoparticles (AuNPs). In the Brust-Schiffrin synthesis, aqueous HAuClX4\ce{HAuCl4} is phase-transferred to toluene by tetraoctylammonium bromide, dodecanethiol is added, and excess NaBHX4\ce{NaBH4} reduces Au(III) to AuNPs of controlled diameter. (4.B-1) Is this top-down or bottom-up? (4.B-3) What is the function of NaBHX4\ce{NaBH4}? (4.B-4) If the average diameter of an AuNP is 3.0 nm and the atomic radius of Au is 0.144 nm, estimate the number of Au atoms per nanoparticle. (4.B-5) Estimate the percentage of Au atoms located at the nanoparticle surface.
Step 4 of 4: Surface atom fraction
Psurface=V(r)−V(r−2rAu)V(r)=1−(1.5−0.2881.5)3=1−0.8083=47%≈40–50%P_{\text{surface}} = \frac{V(r) - V(r - 2r_{\ce{Au}})}{V(r)} = 1 - \left(\frac{1.5 - 0.288}{1.5}\right)^3 = 1 - 0.808^3 = 47\% \approx 40\text{--}50\%
Analysis

The surface atoms form an outer shell of thickness one atomic diameter (2rAu=0.2882r_{\ce{Au}} = 0.288 nm). The shell volume fraction is 1−((1.5−0.288)/1.5)3≈47%1 - ((1.5 - 0.288)/1.5)^3 \approx 47\%, putting the answer in the 40–50 % range (choice b).