The accuracy of a single measurement can be determined by comparison to a "known value". What additional information, if any, is needed in order to state the precision of the measurement?
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Both the Cu2+ and Mg2+ react with the hydroxide ion that i present in producing a blue ppt with the Cu2+ (Cu(OH)2) and a white ppt with the Mg2+ (Mg(OH)2). But with the NH3 present the copper will further react to form Cu(NH3)4+ which dissolves and thus we will have a deep blue solution with the white Mg(OH)2 that can be centrifuged and decanted and thus separated
Describe, using words and chemical equations, the reaction of an aqueous solution of silver nitrate with aqueous hydrochloric acid and the subsequent reaction with aqueous ammonia.Silver nitrate is a soluble salt with the nitrate ion as a spectator ion, the aqueous hydrochloric acid with form a white ppt of AgCl. Then the addition of ammonia will form the silver complex with the NH3 which dissolves the precipitate and forms an aqueous solution. Ag+(aq) + HCl(aq) <---> AgCl(s) + H+ (aq) AgCl(s) + NH3(aq) <---> Ag(NH3)2+(aq) + Cl-(aq)Outline in 5 or so brief steps, the determination of Ka of a weak acid using data from a pH titration.Titrate a known volume of a weak acid with a known concentration of a strong base using a pH meter. Graph the titration curve as pH vs volume of strong base added. Identify the volume of the strong base needed to titrate to the equivalence point. Divide this volume in half and determine from the graph of the pH at this half-equivilance point. At this point in the titration, the pH is equal to the pKA because there are equal amounts of conjugate pairs present in the solution.How would the Ka determined (as in question 9 from above) be affected if you incorrectly over estimated the volume needed to reach the equivilance point of the titration when visually analyzing the titration curve?There is probably minimal impact if the volume is only slightly misread. But too large a volume at the equivalence point results in too large of a calculated volume at the half eqivalence point and a little higher pH value, resulting in a smaller Ka.What is the volume required to reach the half-equivalence point in the titration of a 30.0 mL sample of 0.10 M acetic acid wit a 0.15 M sodium hydroxide solution.30.0 mL x 0.10 M/0.15M=20.0 mL. The volume at the half-equivalence point would be 10.0 mLShow a derivation of the relationship between pH and the pKa at the half-equivalence point,Henderson Hasselbach equation: pH = pKa + log[A]/[HA]. At the half-equivalence point, the [A-] = [HA] and thus pH = pKa = 0 and finally pH = pKa at the half-equivalence point.What is the Ka of a weak acid whose pH at the half-equvialence point is 6.3?pH = pKa = 6.3 thus Ka = 10^-6.3 = 5.0*10^-7In our qualitative analysis labratory, what was the important reason for keeping some of the compound in the hood? (do you know which ones these were?)The concentrated acids/bases are kept in the hood to minimize and/or contain spills and odorsIdentify the particular situation in which you would use a spot test reagent, a precipitating reagent, and a precipitate-dissolving reagentPrecipitating reagents and precipitate-dissolving reagents are used to isolate the ions from a mixture. Spot tests are used to confirm the presence of a suspected ionIn our qualitative analysis scheme, what could be used to separate an aqueous solution of Ni2+ from Ag+?Adding HCl will cause a white ppt of AgCl to form and the Ni2+ ion will not precipitateWhy is it not very useful to use HCl as a spot test reagent between Pb2+ and Fe2+?Fe2+ does not ppt with HCl and PbCl2 is a white solid. There are many white solids and this is not a good confirmation the Pb2+ was presentUsing the indicator chart in your lab manuel identify a range of pH values possible if a solution containing methyl red is yellow and solution containing phenol red is yellowIf methyl red is yellow pH > 6, if phenol red is yellow then pH < 6.6. the pH is between 6 and 6.6List sources of error present in the determination of the enthalpy, entropy, and Gibb's free energy from the precipitation titration and how they would affect the value of the solubility product constant, Ksp.One example, pulling undissolved solid into the pipet when retrieving the dissolved PbCl2 at temperature. This causes too much AgNO3 to be used since there is too much Cl- ion present. Thus calculating too large a concentration of both Cl- and Pb2+ and a calculated Ksp is too largeDescribe the function of the salt bridgeIn a voltaic cell, the salt bridge allows the spontaneous reaction to continue since we do not have a build up of charge in the compartments. The anions flow through the salt bridge into the anode compartment to counter the build up of positive charge and cations flow through the slat bridge into the cathode compartment to counter the build up of negative chargeDetermine the reaction that would occur in a voltaic cell between the two reduction half reactions of Sn2+ going to Sn and Pt2+ going to PtCHanged from Pt2+ to Pb2+ and Pt to Pb. Pb2+(aq) + 2e- <--> Pb(s) Ered = -0.126V Sn2+(aq) + 2e- <--> Sn(s) Ered = -0.136V Turn the tin containin reaction around, Eox = +0.136V and the Ecell = 0.010VA unknown metal is oxidized spontaneously wit Cd2+ being reduced to Cd. The Ecell is measured to be +0.337V. What is the Ered of the unknown metal ion to the elemental metal?Cd2+ to Cd Ered = -0.403V thus 0.337V = -0.403V + Eox, Eox = 0.74VWhat reaction occurs at the cathode in the reaction in problem 24? 25?The reduction half reaction in both cases. In 24, the Pb2+ going to Pb and in 25 the unknown metal cation being reduced to the unknown metal.