14-9. A solution containing 50.0 mL of 0.100 M EDTA buffered to pH 10.00 was titrated with 50.0 mL of 0.020 0 M Hg(CIO4)2 in the cell shown in Exercise 14-B: S.C.E. || titration solution | Hg(1) From the cell voltage E = –0.027 V, find the formation constant of Hg(EDTA)²-.
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- Compute the titration curve for Demonstration 16-1, in which 400.0 mL of 3.75 mM Fe21 are titrated with 20.0 mM MnO24 at a fixed pH of 0.00 in 1 M H2SO4. Calculate the cell voltage at titrant volumes of 1.0, 7.5, 14.0, 15.0, 16.0, and 30.0 mL and sketch the titration curve.A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+ leading to the reaction CH2(COOH)2 + 6Ce4+ + 2H2O ® HCOOH + 2CO2 + 6Ce3+ + 6H+ After standing for 10 minutes at 60oC, the solution was cooled and the x’ss Ce4+ was titrated with 0.1M Fe2+, requiring 14.4 ml to reach the ferroin end point. Calculate the M of the malonic in the sample.For the titration of 100mL of 0.001M NaCl (Cl-) with 0.0100M AgNO3 (Ag+), calculate the voltage at VAg+ = 0.5, 10.0, and 15.0mL Ksp (AgCl) = 1.8 x 10-10.
- A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+ Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions. half‑reaction: ?∘=0.161 V half‑reaction: ?∘=0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. ?E of the Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V. ?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V ?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V ?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V E=0.767…A pH probe/meter uses the following equations: Ecell = L + 0.0592 log a1 = L - 0.0592 pH Where L = L1 + EAg/AgCI + Easy= constants L1 = - 0.0592 log a2 a1 = activity of analyte solution a2 = activity of internal solution Questions: How will measured pH value be affected vs “real” pH if the temperature of the sample is 30C when pH was measured? How will measured pH value be affected vs “real” pH if HCl in pH electrode, became 0.15M instead of 0.1M? How pH value will be affected vs “real” pH if the glass of the pH electrode is not fully hydrated? Please answer all questions and provide a brief explanationA piece of Gold weighing 12,359 Kg is suspected of being contaminated with Iron. To perform an instrumental analysis and To confirm whether or not it contains Fe, a portion of the sample (0.954 g) is taken from the piece and dissolved with 25 mL of aqua regia. Heats up For its complete dissolution, it is cooled and made up to 100 mL. A 10 mL aliquot is taken from this solution and made up to 50 mL. From This last solution is given the appropriate treatment to visualize Fe+2, for which the 1,10-phenanthroline reagent is added. (it forms a complex that is red in color) and is taken to a visible spectrophotometer and with a 12 mm cell a absorbance of 0.45. Previously, a calibration curve of Fe+2 was obtained under the same instrumental conditions obtaining the following data: (view table) Calculate the purity of the gold piece, assuming impurities only due to Fe.
- Consider the titration of 25.0ml 0.0100M Sn2+ by 0.0500M Tl3+ in 1M HCl using Pt and saturated calomel electrodes , calculate E at following volumes of Tl3+ : 1, 5 and 10mlA 50 mL sample solution containing 8-hydroxyquinoline (MW: 145) was analyzed by adding 25 ml, 0.1 M KBrO3, excess KBr and acidified. The mixture was left for 10 minutes in dark place. After this time KI in excess was added followed by titration with 27.9 mL, 0.05 M thiosulfate standard solution. Write balance equations? What is the percent w/v 8-hydroxyquinoline in sample?1. 1093-g sample of impure Na2CO3 was analyzed by residual precipitimetry. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back-titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the endpoint. The percentage Na2CO3 (MW = 106.0 g/mole) in the tested sample is ________ % ? Note: Express final answer using least number of significant figures. 2. The alkalinity of natural waters is usually controlled by OH- (MW = 17.01 g/mole), CO3-2 (MW = 60.01 g/mole), and HCO3- (MW = 61.01 g/mole), which may be present singularly or in combination. Titrating a 10.0-mL sample to a phenolphthalein endpoint requires 38.12 mL of a 0.5812 M solution of HCl, and an additional 18.67 mL of the same titrant to reach the methyl orange endpoint. The composition of the sample is _________% CO3-2 and ___________ % OH- Note: Express final answers using least number of significant figures.
- H2S(aq) is analyzed by titration with coulometrically generated I2 in Reactions 17-3a and 17-3b. To 50.00 mL of unknown H2S sample were added 4 g of KI. Electrolysis required 812 s at 52.6 mA. Find the concentration of H2S (mg/mL)in the sample.A 50.0 mL of 3.1% (w/v, g/mL) NaOH solution is mixed with 40.00 mL of 1.2% (w/v, g/mL) Na2CO3 solution. What volume of 0.087 M HCl will be required to titrate the resultant solution to bromocresol green indicator end point (Bcg pH transition range 3.8-5.4, Fwt of NaOH=40 g/mol, Na2CO3= 105.99 g/mol).Excess Ca(OH)2 is shaken with water to produce a saturatedsolution. The solution is filtered, and a 50.00-mL sampletitrated with HCl requires 11.23 mL of 0.0983 M HCl toreach the end point. Calculate Ksp for Ca(OH)2.
