

Negative Ions
Positive ions will react with negative ions, and vice versa. This is the start of our chemical reactions. For example: Na+ + OH > NaOH (sodium hydroxide) Na+ + Cl > NaCl (salt) 3H^{+} + PO_{4}^{3} > H_{3}PO_{4} (phosphoric acid) 2Na^{+} + S_{2}O_{3}^{2} > Na_{2}S_{2}O_{3} You will see from these examples, that if an ion of one (+), reacts with an ion of one () then the equation is balanced. However, an ion like PO_{4}^{3} (phosphate will require an ion of 3+ or an ion of one (+) (but needs three of these) to neutralise the 3 charge on the phosphate. So, what you are doing is balancing the charges (+) or () to make them zero, or cancel each other out. For example, aluminium exists in its ionic state as Al^{3+}, it will react with many negatively charged ions, examples: Cl^{}, OH^{}, SO_{4}^{2}, PO_{4}^{3}. Let us do these examples, and balance them. Al^{3+} + Cl^{} > AlCl (incorrect) Al^{3+} + 3Cl^{} > AlCl_{3} (correct) How did we work this out?Al^{3+} has three positives (3+) Cl has one negative () It will require 3 negative charges to cancel out the 3 positive charges on the aluminium ( Al^{3+}). When the left hand side of the equation is written, to balance the number of chlorine’s (Cl) required, the number 3 is placed in front of the ion concerned, in this case Cl, becomes 3Cl^{}. On the right hand side of the equation, where the ions have become a compound (a chemical compound), the number is transferred to after the relevant ion, Cl_{3}. Another example:Al^{3+} + SO_{4}^{2} > AlSO_{4} (incorrect) 2Al^{3+} + 3SO_{4}^{2} > Al_{2}(SO_{4})_{3} (correct) Let me give you an easy way of balancing:Al is 3+ SO4 is 2 Simply transpose the number of positives (or negatives) for each ion, to the other ion, by placing this value of one ion, in front of the other ion. That is, Al^{3+} the 3 goes in front of the SO_{4}^{2} as 3SO_{4}^{2}, and SO_{4}^{2}, the 2 goes in front of the Al^{3+} to become 2Al^{3+}. Then on the right hand side of the equation, this same number (now in front of each ion on the left side of the equation), is placed after each “ion” entity. Let us again look at:Al^{3+} + SO_{4}^{2} > AlSO_{4} (incorrect) Al^{3+} + SO_{4}^{2} > Al_{2}(SO_{4})_{3} (correct) Put the three from the Al in front of the SO_{4}^{2} and the 2 from the SO_{4}^{2} in front of the Al^{3+}. Equation becomes: 2Al^{3+} + 3SO_{4}^{2} > Al_{2}(SO_{4})_{3}. You simply place the valency of one ion, as a whole number, in front of the other ion, and vice versa. Remember to encase the SO_{4} in brackets. Why? Because we are dealing with the sulphate ion, SO_{4}^{2}, and it is this ion that is 2 charged (not just the O_{4}), so we have to ensure that the “ion” is bracketed. Now to check, the 2 times 3^{+} = 6^{+}, and 3 times 2^{} = 6^{}. We have equal amounts of positive ions, and equal amounts of negative ions. Another example:NaOH + HCl > ? Na is Na^{+}, OH is OH^{}, so this gave us NaOH. Originally the one positive canceled the one negative HCl is H^{+} + Cl ^{}, this gave us HCl. Reaction is going to be the Na^{+} reacting with a negatively charged ion. This will have to be the chlorine, Cl^{}, because at the moment the Na^{+} is tied to the OH^{}. So: Na^{+} + Cl^{} > NaCl The H+ from the HCl will react with a negative () ion this will be the OH^{} from the NaOH. So: H^{+} + OH^{} > H_{2}O (water). The complete reaction can be written:NaOH + HCl > NaCl + H_{2}O. We have equal amounts of all atoms each side of the equation, so the equation is balanced. orNa^{+}OH^{} + H^{+}Cl^{} > Na^{+}Cl^{} + H^{+}OH^{} Something More Difficult:Mg(OH)_{2} + H_{3}PO_{4} > ? (equation on left not balanced) Mg^{2+} 2OH^{} + 3H^{+}PO_{4}^{3} > ? (equation on left not balanced), so let us rewrite the equation in ionic form. The Mg^{2+} needs to react with a negatively charged ion,
this will be the PO_{4}^{3}, (Remember the swapping of the positive or negative charges on the ions in the left side of the equation, and placing it in front of each ion, and then placing this number after each ion on the right side of the equation) What is left is the H^{+} from the H_{3}PO_{4} and this will react with a negative ion, we only have the OH^{} from the Mg(OH)_{2} left for it to react with. 6H^{+} + 6OH^{} > 6H_{2}O Where did I get the 6 from? W hen I balanced the Mg^{2+} with the PO_{4}^{3}, the equation became 3Mg^{2+} + 2PO_{4}^{3} > Mg_{3}(PO_{4})_{2} Therefore, I must have required 3Mg(OH)_{2} to begin with, and 2H_{3}PO_{4}, ( because we originally had (OH)_{2} attached to the Mg, and H_{3} attached to the PO_{4}. I therefore have 2H_{3} reacting with 3(OH)_{2}. We have to write this, on the left side of the equation, as 6H^{+} + 6OH^{} because we need it in ionic form. The equation becomes: 6H^{+} + 6OH^{} > 6H_{2}O The full equation is now balanced and is: 3Mg(OH)_{2} + 2H_{3}PO_{4} > Mg_{3}(PO_{4})_{2} + 6H_{2}O I have purposely split the equation into segments of reactions. This is showing you which ions are reacting with each other. Once you get the idea of equations you will not need this step. The balancing of equations is simple. You need to learn the valency of the common ions (see tables). The rest is pure mathematics, you are balancing valency charges, positives versus negatives. You have to have the same number of negatives, or positives, each side of the equation, and the same number of ions or atoms each side of the equation. If one ion, example Al^{3+}, (3 positive charges) reacts with another ion, example OH^{} (one negative ion) then we require 2 more negatively charged ions (in this case OH^{}) to counteract the 3 positive charges the Al^{3+} contains. Take my earlier hint, place the 3 from the Al^{3+} in front of the OH^{}, now reads 3OH^{}, place the 1 from the hydroxyl OH^{} in front of the Al^{3+}, now stays the same, Al^{3+} (the 1 is never written in chemistry equations). Al^{3+} + 3OH^{} > Al(OH)_{3} The 3 is simply written in front of the OH^{}, a recognised ion, there are no brackets placed around the OH^{}. On the right hand side of the equation, all numbers in front of each ion on the left hand side of the equation are placed after each same ion on the right side of the equation. Brackets are used in the right side of the equation because the result is a compound. Brackets are also used for compounds (reactants) in the left side of equations, as in 3Mg(OH)_{2} + 2H_{3}PO_{4} > ?
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The following page shows one of many emails I have received basically asking the same question, so I have posted the answer here... More Equations

