Barium Reacts With A Polyatomic Ion To: Fill & Download for Free

Note down the ions that are present in the reaction. They are:Sodium Ion (Na[math]^+[/math])Phosphate (PO[math]_4[/math][math]^{3-}[/math])Barium Ion (Ba[math]^{2+}[/math])Chloride (Cl[math]^-[/math])Since Barium ion has a charge of 2+, then we need two Chloride atoms since they have a charge of 1- to make a neutral compound. So, the formula would be BaCl[math]_2[/math].Similarly, the formula for Sodium Phosphate would be Na[math]_3[/math]PO[math]_4[/math].Now, in Barium Phosphate Barium ion has a charge of 2+ and Phosphate a charge of 3-. To form a neutral compound, we need to take the LCM of charges (that would be 6) hence, we need three Barium ions and two phosphate ions. This would give us a compound with the formula Ba[math]_3[/math](PO[math]_4[/math])[math]_2[/math].Similarly, in Sodium Chloride the Sodium ion has a charge of 1+ while the chloride ion has a charge of 1-. Since they are already balanced, the compound has a formula NaCl.The left hand side (Reactants) looks like:BaCl[math]_2[/math] + Na[math]_3[/math]PO[math]_4[/math]The right hand side (Products) looks like:Ba[math]_3[/math](PO[math]_4[/math])[math]_2[/math] + NaClNow comes the balancing part. Always balance the ones with greater magnitude of charges first.The Reactant side has only one phosphate ion but the Product side has 2. So, we add a coefficient 2 in front of Na[math]_3[/math]PO[math]_4[/math] since it is the source of the phosphate ion. This also has another consequence of giving rise to a total of 6 instead of previous 3 sodium ions.The Reactant side has only one Barium atom but the Product side has 3. To balance, we add '3' coefficient to BaCl[math]_2[/math] since it gives the Barium ion. This also has another consequence of giving rise to a total of 6 instead of previous 2 chloride ions.From the above results, we now have 6 Sodium and Chloride ions which together give NaCl. The equation is balanced.Therefore, the equation will be:3 BaCl[math]_2[/math] + 2 Na[math]_3[/math]PO[math]_4[/math] [math]\rightarrow[/math] Ba[math]_3[/math](PO[math]_4[/math])[math]_2[/math] + 6 NaCl.You could also use linear algebra to balance it. This is especially good for balancing tough ones. Just see the number of ion in question to get for a linear expression. Assume a coefficient for every species on both sides. For example,a CH[math]_3[/math]OH + b O[math]_2[/math] [math]\rightarrow[/math] c CO[math]_2[/math] + d H[math]_2[/math]ONow,There are a*1 carbon atoms on the left hand side while there are c*1 carbon atoms on the right hand side. So,a = cThere are a*4 hydrogen atoms on the left hand side while there are d*2 hydrogen atoms on the right hand side. So,4a = 2dThere are (a*1 + b*2) oxygen atoms on the left hand side while there are (c*2 + d*1) oxygen atoms on the right hand side. So,a + 2b = 2c + dNow, substitute a value in any of the variables to get the other values. For example, if I consider a = 1, then I get the values:c = 1d = 2b = 3/2If you do not like fractions then you can multiply by the fraction's denominator to get:a = c = 2d = 4b = 3You get:2 CH[math]_3[/math]OH + 3 O[math]_2[/math] [math]\rightarrow[/math] 2 CO[math]_2[/math] + 4 H[math]_2[/math]O.For polyatomic ions, it is easier to consider it as a single entity (you could substitute it with a big X). For example,w BaCl[math]_2[/math] + x Na[math]_3[/math]PO[math]_4[/math] [math]\rightarrow[/math] y Ba[math]_3[/math](PO[math]_4[/math])[math]_2[/math] + z NaClorw BaCl[math]_2[/math] + x Na[math]_3[/math]X [math]\rightarrow[/math] y Ba[math]_3[/math]X[math]_2[/math] + z NaClFor Barium,w = 3yFor Sodium,3x = zFor Chloride,2w = zFor Phosphate (X),x = 2yNow, simply substitute any value (usually 1) to any variable to get the values. For example, if I consider y=1, then I get:x = 2w= 3z = 6

A hydroxide refers to the OH- polyatomic ion and is formed when an oxygen makes a covalent bond with one hydrogen (however you would not see such ions free in nature as they would more probably be in compounds).Potassium hydroxide (KOH) is formed when Potassium forms ionic bonds with OH- ions while Potassium Oxide (K2O) is formed when potassium forms ionic bonds with the Oxide (O2-) ions.Hydrochloric acid + Potassium Hydroxide ---> Potassium Chloride + Wateri.e. HCl(aq) + KOH (aq) ----> KCl (aq) + H2O (l)This reaction is a neutralization reaction and occurs when an acid (HCl) reacts with a base (KOH).Potassium hydroxide KOH is a colorless, odorless, corrosive, deliquescent crystalline solid. It readily absorbs water and carbon dioxide from air. KOH has a good solubility in water, 49.4% wt at 0°C, its solvation is highly exothermic. It also dissolves in methyl alcohol (35.5% wt at 28°C), ethanol (27.9% wt at 28°C). Potassium hydroxide forms stable mono-, di- and tetrahydrates (see table below):Anhydrous potassium hydroxide exists in two crystalline modifications. Monoclinic modification is stable below 247°C, at higher temperature KOH has a NaCl-like cube lattice.Potassium hydroxide is a "strong base", along with other alkalis such as sodium hydroxide, lithium hydroxide, calcium hydroxide. Solid KOH violently reacts with mineral acids (neutralization reaction) to form salts. It also readily combines with their anhydrides (acidic oxides SO2, CO2, NO2and others) to give corresponding salts. Thus, interaction with carbon dioxide results in potassium hydrocarbonate:KOH + CO2 → KHCO3It also easily reacts with other "acidic" gases such as hydrogen chloride HCl, hydrogen bromide HBr, dihydrogen sulfide H2S to produce salts:KOH + HCl → KCl + H2OInteraction of KOH with hydrogen fluoride leads to a mixture of potassium fluoride KF, potassium bifluoride KF*HF and potassium trifluoride KF*2HF. It reacts with carbon monooxide to form potassium formate:KOH + CO → HC(O)OKAnhydrous KOH combines with chlorine or bromine only if temperature higher 600°C.Molten potassium hydroxide reacts with aluminum, zinc, gallium, beryllium, tin, lead and antimony releasing dihydrogen H2 and to give oxometallates such as KGaO2, KSnO2and so on. Same products may be obtained by the reaction of KOH with these metals oxides and hydroxides.Potassium hydroxide is fully ionized (dissociated) in water solutions. Its water solution dissolves silicon, germanium, boron, their oxides and acids to give corresponding salts. It lixiviates silicate glasses converting SiO2 into potassium polysilicates. After this action glass acquires ulcer, rough surface and becomes fragile.Potassium hydroxide forms compounds and solid solutions with other alkali and alkali-earth metal hydroxides and salts. Thus, addict with lithium hydroxide has a formula KOH*2LiOH (m.p. 313°C, decomposes). With NaOH it forms solid solutions and eutectic (m.p. 170°C, 50% mol of KOH). Continuous series of solid solutions is formed with rubidium hydroxide RbOH. With barium hydroxide KOH forms solid solutions and adduct KOH*4Ba(OH)2 (m.p. 390°C). It also forms eutectic with potassium carbonate K2CO3 (m.p. 360°C, 78% wt of KOH).Potassium hydroxide reacts with a number of organic compounds. Thus, interaction with carboxylic acids, their anhydrides, amides results in corresponding potassium salts. Alkyl- and arylhalogenated derivatives being treated with KOH convert into alcohols. It also saponifies esters to produce carboxylic acid salt and alcohol:R1C(O)OR2 + KOH → R1COOK + R2OHPotassium hydroxide, either in anhydrous or dissolved form, can pose several risks. It may cause chemical burns of skin or mucous membranes, permanent injury or scarring, and blindness. It is able to permeate deep into animal tissue damaging adipose, lipids and proteins

The answer depends on whether you want to distinguish them based on observable properties or make an educated guess based on name and chemical formula.Introductory level chemistry (ie; high school level) provides a few generalizations that can help students make a decent educated guess as to the nature of the compound.Lets look at chemical formula/names first.If the compound contains a metal element, it is most likely going to be ionic. Simple binary compounds consisting of a metal element and a non-metal element tend to be ionic (ex. Sodium chloride NaCl, iron (III) bromide FeBr3, magnesium nitride Mg3N2). When metals and non-metals react, they usually create ions as the valence electrons from the metal are lost (forming a metal ion with a positive charge) while the non-metal gains these electrons (forming a negative ion).Things get a little more tricky when polyatomic ions are introduced. Polyatomic ions are formed when a group of atoms bound together by covalent bonds also have an excess or deficiency of electrons, resulting in the particle having a charge. These include ions such as carbonate ([CO3]2-), nitrate ([NO3]1-), sulfate ([SO4]2-), phosphate ([PO4]3-), ammonium ([NH4]1+) and many more. However, except for ammonium, most of these ions are still usually found associated with a metal ion (ex, calcium nitrate Ca(NO3)2, zinc phosphate (Zn3(PO4)2), silver sulfate (Ag2SO4), barium carbonate BaCO3). Ammonium, being a positively charged ion will be associated with a non-metal or another polyatomic ion (ammonium chloride NH4Cl, ammonium carbonate (NH4)2CO3).Also, molecular compounds tend to be named differently than ionic compounds. These compounds names will often use a prefix system (mono, di, tri…) that tells you the number of atoms of each element in one molecule of the compound (ex, carbon dioxide CO2, carbon monoxide CO, dinitrogen tetroxide N2O4). You don't see these prefixes used in the names of ionic compounds.Ionic and molecular compounds differ greatly in their physical properties, so observations can also be used to distinguish them from each other.Ionic compounds are generally crystalline solids at room temperature and have very high melting points (ie; > 300 C). Molecular compounds, on the other hand, may be gases (ie; carbon dioxide), liquids (ie; water) or solids (ie; sugar) at room temperature. Even when solid, however, they tend to have lower melting points (<300 C). So, if a compound is a gas or liquid at room temperature, it's most likely a molecular compound. If it's a solid, check melting points.Ionic compounds tend to be water soluble (although their solubility varies) and when dissolved produce solutions that conduct electricity (ie; they are electrolytes). Molecular compounds may dissolve in water (if they are very polar) or they may be insoluble in water (if they are non-polar). However, even if they dissolve in water, they tend to be non-electrolytes, so the solution will not conduct electricity. (Unfortunately, there are a few exceptions, some molecular compounds ionize in water, specifically, those that become acids when added to water, like HCl, so these solutions will conduct electricity, but the original compound was a gas).So, if you had a sample of table salt (ionic) and sugar (molecular), both appear as white, crystalline solids. So just looking at them you may not be able to tell them apart. But, you could perform a few simple tests on them.Try to melt them. Sugar can easily melt if placed in a spoon and held over an burning candle, but salt will not.Try to dissolve them in water. Both dissolve quite well in water. But, a conductivity test will show that the salt solution is a good electrolyte while the sugar solution is not.