1 Colloid Chemistry in Mineral Processing 1. 1. Work of adhesion and work of cohesion The work of adhesion between two phases (for example, between two immiscible liquids, or between solid and liquid) is equal to the reversible work required Wu=2YLv=-△G yur Yaw WSL=YSV+yLV-YSL=-△G yu ysr Fig. 1.1 to separate unit area of the liquid-liquid interface(for a two phase liquid-liquid system) and form two separate liquid air interfaces(see Fig. 1.1)and is given by the Dupre equation Wa=Ya+Yb-Yab (1.1) where ya and yb are the surface tensions of phases a and b, respectively, and Yab is the interfacial tension a-b. The work of cohesion for a single liquid corresponds to the work required to pull apart a column of liquid of unit cross-sectional area and is given by W2=2y。 (1.2) 1.2. Wettability In the case of a solid wetted by a liquid, the wettability is determined by the solid- liquid work of adhesion(WsL) and the work of cohesion of liquid (WLL). The classical boundary condition for the hydrophilic-hydrophobic transition is equality of Wst and WLL. According to Dupre's equation (Eq. 1.1)， for a solid-liquid system Wst is given by WSL=YSv+YLV-YSL (1.3) and
1 Colloid Colloid Colloid Colloid Chemistry Chemistry Chemistry Chemistry in Mineral Mineral Mineral Mineral Processing Processing Processing Processing 1. 1. Work of adhesion adhesion adhesion adhesion and work of cohesion cohesion cohesion cohesion The work of adhesion between two phases (for example, between two immiscible liquids, or between solid and liquid) is equal to the reversible work required Fig. 1.1 to separate unit area of the liquid-liquid interface (for a two phase liquid-liquid system) and form two separate liquid air interfaces (see Fig. 1.1) and is given by the Dupre equation Wa a b ab = γ + γ − γ (1.1) where γa and γb are the surface tensions of phases a and b, respectively, and γab is the interfacial tension a-b. The work of cohesion for a single liquid corresponds to the work required to pull apart a column of liquid of unit cross-sectional area and is given by Wc a = 2γ (1.2) 1.2. Wettability Wettability Wettability Wettability In the case of a solid wetted by a liquid, the wettability is determined by the solidliquid work of adhesion (WSL) and the work of cohesion of liquid (WLL). The classical boundary condition for the hydrophilic-hydrophobic transition is equality of WSL and WLL. According to Dupre’s equation (Eq. 1.1), for a solid-liquid system WSL is given by WSL SV LV SL = γ + γ − γ (1.3) and
2(1.4)Wu=2LvSince there are no experimental methods to determine the surface tension of asolid (sv and s) equation 1.3 can only be solved with the use of Young's equation(Fig, 1.2)(1.5)YSV=SL+YLVCOSOYtvYsvFig. 1.2. Sessile drop on a smooth solid surfaceBy combining Eqs. 1.3. and 1.5 one can get(1.6)Wst=Lr(cosO+I)Since both liquid surface tension (yv)and the contact angle ()can experimentally bedetermined, equation 1.6 is often used.Equations 1.4 and 1.6giveWa - r(I+cos @)(1.7)Wu2LVandWsL.(1.8)cOsO=2WuThe condition of hydrophobicity follows from Eq. 1.8; only for Ws.<Wu①+0.Fig. 1.3 depicts a sessile droplet resting on a flat solid surface; as this figureshows, the contact angle at the solid/liquid interface is determined by the values of theCOHESIONGasLiquidADHESIONSolid
2 WLL LV = 2γ (1.4) Since there are no experimental methods to determine the surface tension of a solid (γSV and γSL) equation 1.3 can only be solved with the use of Young’s equation (Fig. 1.2) γ SV = γ SL + γ LV cosΘ (1.5) Fig. 1.2. Sessile drop on a smooth solid surface. By combining Eqs. 1.3. and 1.5 one can get = (cosΘ + 1) WSL LV γ (1.6) Since both liquid surface tension (γLV) and the contact angle (Θ) can experimentally be determined, equation 1.6 is often used. Equations 1.4 and 1.6 give LV LV LL SL W W γ γ 2 (1+ cos Θ) = (1.7) and cosΘ = 2 −1 LL SL W W (1.8) The condition of hydrophobicity follows from Eq. 1.8; only for WSL <WLL Θ ≠ 0. Fig. 1.3 depicts a sessile droplet resting on a flat solid surface; as this figure shows, the contact angle at the solid/liquid interface is determined by the values of the
3work of adhesion of the liquid to the solid, and the work of cohesion of the liquidTheworkofadhesionofliquidtosolid(Wsi)canbesplitintoseveralcontributions (Fowkes). For water molecular interacting with a solid, the most importantare:van der Waals dispersion forces contribution (Ws),hydrogen bond contribution(Wsth)and, if the solid has an electrical charge,an electrical forces contribution (Ws)ThenWst = Ws + W +Ws(1.9)For the solid which does not have any polar groups and, therefore, does not interact withwater molecules through hydrogen bonding ( wst ~ O), and does not have an electricalcharge so that W ~0, substitution of Eq 1.9 to Eq. 1.8 giveswi(1.10)cOsO~21WuOMooSoCFig.1.4.Graphite and molybdenitecrystal latticesUsing this equation, Laskowski and Kitchener (1969) showed that because ofexceptionallyhighworkof cohesionofwater(Wu=145mJ/m2)forall solidsWd<Wu;therefore, all solidswould behydrophobic (@>O)if they interacted withwater only through dispersion forces().This explains why some minerals (such asgraphite,molybdenite, sulfur, talc)are hydrophobic by nature since. This is perfectly in line with Gaudin who pointed out that all hydrophobic solids presentnon-polar molecular groups whereas hydrophilic solids have ionic or dipolar groupscapable of undergoing hydrationBy writing Eq. 1.8 in the formWs+Ws+Ws-1COsO=2(1.11)Wu
3 work of adhesion of the liquid to the solid, and the work of cohesion of the liquid. The work of adhesion of liquid to solid (WSL) can be split into several contributions (Fowkes). For water molecular interacting with a solid, the most important are: van der Waals dispersion forces contribution (WSL d ), hydrogen bond contribution (WSL h ) and, if the solid has an electrical charge, an electrical forces contribution (WSL e ). Then e SL h SL d WSL =WSL +W +W (1.9) For the solid which does not have any polar groups and, therefore, does not interact with water molecules through hydrogen bonding ( ≈ 0 h WSL ), and does not have an electrical charge so that ≈ 0 e WSL , substitution of Eq 1.9 to Eq. 1.8 gives cosΘ ≈ 2 −1 LL d SL W W (1.10) Fig. 1.4. Graphite and molybdenite crystal lattices. Using this equation, Laskowski and Kitchener (1969) showed that because of exceptionally high work of cohesion of water (WLL = 145 mJ/m2 ) for all solids LL d WSL <W ; therefore, all solids would be hydrophobic ( Θ > 0) if they interacted with water only through dispersion forces( ). This explains why some minerals (such as graphite, molybdenite, sulfur, talc) are hydrophobic by nature since . This is perfectly in line with Gaudin who pointed out that all hydrophobic solids present non-polar molecular groups whereas hydrophilic solids have ionic or dipolar groups capable of undergoing hydration. By writing Eq. 1.8 in the form cos 2 −1 + + Θ = LL e SL h SL d SL W W W W (1.11)
4it becomes obvious that solid surface properties are strongly affected by the electricalsurface charge.1.3.Origin of electrical charge at solid/liquid interface(A). Ionic Solids. In the case of ionic solids such as Agl, BasO4, CaF2 etc. (soluble saltsalso belong to this group but will be discussed separately), the surface charge arises fromthetransportacrosstheinterfaceof ionsconstitutingthelattice.Thoseparticularionsthatare free to pass between both phases and therefore establish the electrical charge arecalledpotential-determiningions(PDI).ForAgthepotential-determiningionsareAgtand I,for a solid like calcite, CaCOs, the potential-determining ions are Ca2+ and CO,2but since their concentration also depends on pH, the potential-determining ions for thismineral are Ca 2+, CO, 2- and also Ht, OH and HCO3Agl/aqueous solutioninterphaseFig.1.5.Agl/aqueous solution interfaceThe surface charge of a solid is determined by theconcentration of potential-determiningionsonthesolidsurface(orinotherwordsisdeterminedbyadsorptiondensityofpotential-determining ions). In case of Agl surface, surface charge is given by(1.12)0, = F(T+-F-)where FAg+ is the adsorption density in mol/cm? of Agt ions, and Fi is that ofI ion, and Fis theFaraday constant.Agl is practically insoluble in water. The solubility product of AglK,sp = [Ag [/-] = 10-16(1.13)It was experimentally determined that in the system in which Agl is in equlibriumwith aqueous solution, the Agl/water interface does not have electrical charge at Agt ionsconcentration [Ag*] = 10-5.5 mol/dm3. In analogy with pH (pH = -log [H*D), suchconcentrations are expressed as pAg+=5.5. (pAg*=-log [Ag*1)
4 it becomes obvious that solid surface properties are strongly affected by the electrical surface charge. 1.3. Origin of electrical electrical electrical electrical charge at solid/liquid solid/liquid solid/liquid solid/liquid interface interface interface interface (A). Ionic Solids. In the case of ionic solids such as AgI, BaSO4, CaF2 etc. (soluble salts also belong to this group but will be discussed separately), the surface charge arises from the transport across the interface of ions constituting the lattice. Those particular ions that are free to pass between both phases and therefore establish the electrical charge are called potential-determining ions (PDI). For AgI, the potential-determining ions are Ag+ and I - , for a solid like calcite, CaCO3, the potential-determining ions are Ca 2+ and CO3 2- but since their concentration also depends on pH, the potential-determining ions for this mineral are Ca 2+ , CO3 2- and also H+ , OHand HCO3 - . Fig. 1.5. AgI/aqueous solution interface The surface charge of a solid is determined by the concentration of potential-determining ions on the solid surface (or in other words is determined by adsorption density of potential-determining ions). In case of AgI surface, surface charge is given by = (Γ + − Γ − ) Ag I σ s F (1.12) where ΓAg+ is the adsorption density in mol/cm2 of Ag+ ions, and ΓI- is that of I - ion, and F is the Faraday constant. AgI is practically insoluble in water. The solubility product of AgI 16 [ ][ ] 10 + − − KSP = Ag I = (1.13) It was experimentally determined that in the system in which AgI is in equlibrium with aqueous solution, the AgI/water interface does not have electrical charge at Ag+ ions concentration [Ag+ ] = 10-5.5 mol/dm3 . In analogy with pH (pH = -log [H+ ]), such concentrations are expressed as pAg+ = 5.5. (pAg+ = - log [Ag+ ])
5Theprevious equations is commonly written(1.14)pAgt + pI = 16(in the same way for H+ and OH ions in water, pH + pOH= 14)So, the point-of-zero-charge (p.z.c.) conditions for Agl crystals, which are inequilibrium with aqueous solution, are reached when the concentrations of potential-determining ions are: pAg*=5.5 (and pl = 10.5, since pAg+pl =16). In other words,at p.z.c.the concentration of [Ag*] =10-5.5.mol/dm3 is 10s times higher than theconcentration ofl ions (since at pzc pl = 10-10.5 mol/dm)For ionic solids the surfacepotential, o,can be calculated from the equationCRT(1.15)PZFCrFor the case of Agl this equation at room temperature givesY。= 0.059(pAg*o - pAgt) = 0.059(pl- pL) Volt(1.16)so to sum it up, particles of Agl in aqueous solutions are not electrically charged onlywhen [Ag'] = 10-5.5 (thus when []= 10-10.5). This shows that the concentration of silverionsinthesolutionmustbe1ostimeshigherthantheconcentrationof iodideionsfortheAgl particle not to have an electrical charge. This results from much higher hydrationenergy of Agions in comparison with that of I ions.+118459ASV42AA办814131218151067pl
5 The previous equations is commonly written pAg+ + pI - = 16 (1.14) (in the same way for H+ and OHions in water, pH + pOH = 14) So, the point-of-zero-charge (p.z.c.) conditions for AgI crystals, which are in equilibrium with aqueous solution, are reached when the concentrations of potentialdetermining ions are: pAg+ = 5.5 (and pI - = 10.5, since pAg+ + pI - = 16). In other words, at p.z.c. the concentration of [Ag+ ] = 10-5.5 . mol/dm3 is 105 times higher than the concentration of I - ions (since at pzc pI - = 10-10.5 mol/dm3 ). For ionic solids the surface potential, ψo , can be calculated from the equation o M M o C C zF RT + + Ψ = ln (1.15) For the case of AgI this equation at room temperature gives 0.059( ) 0.059( ) o o o pAg pAg pI pI + + − − Ψ = − = − Volt (1.16) so to sum it up, particles of AgI in aqueous solutions are not electrically charged only when [Ag+ ] = 10-5.5 (thus when [I - ] = 10-10.5 ). This shows that the concentration of silver ions in the solution must be 105 times higher than the concentration of iodide ions for the AgI particle not to have an electrical charge. This results from much higher hydration energy of Ag+ ions in comparison with that of I - ions