Why is it more difficult to identify all of ions of a mixture than to identify the cation and anion in a binary ionic compound?
It is easier to separate two ions with different physical properties, such as Na+ and Cl- . When trying to identify components in a mixture, there is a wide range of physical properties one has to identify and associate to a very specific particle, when several cationic and anionic groups have very similar properties (for example: Li+, Na+ and K+, or Mg2+ and Ca2+)
Many metal ions are precipiated from solution by the sulfide ion.?
find moles 0.0275 L of 0.121 mol/litre CuSO4 = 0.0033275 moles CuSO4 by the equation CuSO4(aq) + Na2S)aq)--> CuS(s)+ Na2SO4(aq) 0.0033275 moles CuSO4 reacts with an equal number of moles of Na2S = 0.0033275 moles Na2S find volume 0.0033275 moles Na2S @ 0.105 mol/litre Na2S = 0.03169 litres that's 31.69 ml of the 0.105M Na2S solution your answer rounded to 3 sig figs is 31.7 ml of the Na2S solution
How are anions and cations formed?
POSITIVE AND NEGATIVE IONS: CATIONS AND ANIONSCations (positively-charged ions) and anions (negatively-charged ions) are formed when a metal loses electrons, and a nonmetal gains those electrons. The electrostatic attraction between the positives and negatives brings the particles together and creates an ionic compound, such as sodium chloride.A metal reacts with a nonmetal to form an ionic bond. You can often determine the charge an ion normally has by the element’s position on the periodic table:The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.The halogens (VIIA elements) all have seven valence electrons. All the halogens gain a single electron to fill their valence energy level. And all of them form an anion with a single negative charge.The VIA elements gain two electrons to form anions with a 2- charge.The VA elements gain three electrons to form anions with a 3- charge.The first table hows the family, element, and ion name for some common monoatomic (one atom) cations. The second table gives the same information for some common monoatomic anions.Some Common Monoatomic CationsFamilyElementIon NameIALithiumLithium cationSodiumSodium cationPotassiumPotassium cationIIABerylliumBeryllium cationMagnesiumMagnesium cationCalciumCalcium cationStrontiumStrontium cationBariumBarium cationIBSilverSilver cationIIBZincZinc cationIIIAAluminumAluminum cationSome Common Monoatomic AnionsFamilyElementIon NameVANitrogenNitride anionPhosphorusPhosphide anionVIAOxygenOxide anionSulfurSulfide anionVIIAFluorineFluoride anionChlorineChloride anionBromineBromide anionIodineIodide anionIt’s more difficult to determine the number of electrons that members of the transition metals (the B families) lose. In fact, many of these elements lose a varying number of electrons so that they form two or more cations with different charges.The electrical charge that an atom achieves is sometimes called its oxidation state. Many of the transition metal ions have varying oxidation states. The next table shows some common transition metals that have more than one oxidation state.Some Common Metals with More than One Oxidation State
Calculate the [OH-] and pH of the solution?
Calculate [OH-] and pH of a 0.10 M NaCN solution. HCN (Ka = 4.9*10^(-10)) Do I write the equation as follows: NaCN(aq) → Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound (composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water, therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) → HCN(aq) + OH-(aq) Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ .......... 0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb = 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2 Is this solution correct? I tried doing it myself, but I'm pretty unsure of my work, so I would appreciate if someone could verify it. I have a test on Monday, but it would be nice if I could get immediate feedback on how I'm doing.
How are different ionic compounds distinguished from a solution if their ions are dissociated and all mixed up?
When you have two kinds of salts dissolved in water, the anions and the cations will be scrambled. If you strip off the water, you might recover four different salts instead of the two you put in. But there are methods to separate and identify the two cations and the two anions that were present originally.Anion exchange resins can be used to chromatographically separate anions. For example, the Cl- , NO2- - , PO4 - - - , Br - , and SO4- - anions will elute through the resin at different speeds, and can be identified and quantitated.If that same aqueous solution of the two salts is passed through a cation exchange resin, you could do the same for the Na+ , Co++, Fe+++, K+ and Sn++ cations.There could be a problem with the H+ and OH - ions originally present, aside from the difficulty of actually doing the analyses itself.
Arrange the ions N3-, O2-, Mg2+, Na+, and F- in order of increasing ionic radius...?
The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. Mg2+ Na+ F- O2- N3-