Unit 1 And 4 Reaction Lab Report

Conclusion:
In Station 1- reaction 2, Station 3- reaction 1, Station 4-reaction 1, Station 5 reaction 1, Station 6-reaction 1 and 2, and Station 7- 1 and 2 there was chemical reaction because all of those mixed ionic compounds created a precipitate. However some solutions did not become insoluble and produce precipitates such as: Station 1- BaCl2 (aq) + KNO3 (aq) , Station 2- KNO3(aq) + AgNO3(aq) and KNO3(aq) + CaSO4(aq), Station 3- Na2CO3(aq) + KNO3(aq) , Station 4- NaPO4(aq) + KNO3(aq), and Station 5- 2Na3PO4(aq) + KNO3(aq). The lab demonstrated the Solubility Rules when the equations that included Nitrates and Group 1 salts resulted in soluble compounds, and the ionic compounds that included Carbonates, Phosphates, Hydroxides,

They determine whether the aqueous solutions produce soluble or insoluble products. If the compounds fit the requirements of an soluble reaction it was soluble and vise-versa. For example, one of the products in Station 2 was Potassium Nitrate (KNO3 ), and this product was soluble due to the solubility rules that state: all Nitrates are soluble. Another example is shown in Station 3- reaction 1, which had NaCO3 (aq) and CaSO4 (aq) as the reactants to produce an insoluble product CaCO3(s). That proves the solubility rule that all Carbonates except Group 1A are insoluble.Solutions with the mixed volumes of .10 M

This lab only included double-replacement reaction which allowed for only one of 2 types of products. Products that chemically reacted (solid) or products that didn’t (aqueous). The insoluble products of these double replacement reaction occurred when the cations or anions of the reactant bonded with the cation or anion of the other reaction.When this happens the reactants get paired together with the reactant it bonded with and causes a replacement. This is shown evident in the lab in Station 4- reaction 1 with the reactants Na3PO4(aq) AgNO3 (aq) → Ag3PO4 (s) and NaNO3(aq). Ag+1 and PO-34 produce an insoluble product due to Silver’s +1 charge and Phosphate’s -3 charge. For ex, if both ions had the same charge , the ions wouldn’t be attracted to each other. The ions then wouldn’t be able to form a strong ionic bond and the solution would remain transparent, clear, and soluble. IMFs also play a major part in solubility and the state of which the substance is in. Since intermolecular forces determine how strong or weak the attraction the particles are, the substance could be a gas, liquid, or solid. If the