Q: A. Please provide the products of each reaction. H₂N CI CI OH F NaOCI CH3CN / Acetic Acid
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A: Given: VPb(NO3)2=175mL;[Pb(NO3)2]=0.270M;[S2−]=0.0263M%Pb=???Step 1: Write the balanced…
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- 0.0322 M EDTA solution is used as the titrant in the titration of a 50.00 mL solution that is 0.0129 M in Cu2+. The solution is held at pH = 10.00 by a buffer that is 0.050 in ammonia (NH3). At what volume (in mL) of EDTA added will the equivalence point be reached?Calculate the pNi2+ at the equivalence point for the titration of a 20.00 mL solution of 0.0500 M NaOH with 0.045 M NiCl2 to precipitate Ni(OH)2 (Ksp = 6 x 10-16).Explain why K2CrO4 is used as an indicator in Argentometric titration?
- will H3N titrate with 1.0M Ba(OH)2?In this experiment, the color change from pink/violet to colorless during the titration is caused by: change from MnO4– to Mn2+ change from Fe2+ to Fe3+ change from H+ to H2O the addition of an indicator.Average [EDTA] = C: Determination of the hardness of an unknown water sample: Using an unknown water sample, three 25 mL samples in 3 separate 250 mL Erlenmeyer flasks are prepared. To each flask, 5 mL of pH pH10 buffer, 3 to 5 drops of Eriochrome Black T indicator, and 15 drops of 0.03 M MgCl2 solution are added. Each flask is titrated to the end point using EDTA.
- A 0.4126-g sample of primary-standard Na₂CO₃ was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove CO₂, following which the excess HClO₄ was back-titrated with 9.20 mL of dilute NaOH. In a separate experiment, it was established that 26.93 mL of the HClO₄ neutralized the NaOH in a 25.00-mL portion. Calculate the molarities of the HClO₄ and NaOH. Input your answer in compound form (i.e. x.xx M HClO4 and x.xx M NaOH) with the correct significant figures.A 0.4126-g sample of primary-standard Na2CO3 was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove CO2, following which the excess HClO4 was back - titrated with 9.20 mL of dilute NaOH. In a separate experiment, it was established that 26.93 mL of the HClO4 neutralized the NaOH in a 25.00-mL portion. Calculate the molarities of the HClO4 and NaOH.A 1.43 x 10-4 L solution of 45.7 mg/mL protein was analyzed using the Kjeldahl procedure. After digestion of the protein, the liberated NH3 was collected in 9.80 mL of 0235 M HCl. The unreacted acid required 10.75 mL of a standard solution of 0.0149 M NaOH for complete titration. Calculate the weight percent of nitrogen in the protein. (Molar mass N = 14.00674 g/mol)
- 25.00 mL of a saturated Ca(OH)2 sample requires 22.50 mL of 0.0250 M HCl to neutralize it. Calculate the value for Ksp of Ca(OH)2 from this data. Given the equilibrium equation: Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH– (aq) 1) Write the Ksp expression? 2) Use titration data to determine moles of OH¯ in the 25.0 mL sample (Remember, every one H+ neutralizes one OH¯.) 3) Use moles of OH¯ and sample volume to determine [OH¯]: 4) Determine [Ca2+]: 5) Calculate the Ksp for Ca(OH)2Compare Tris and PBS as buffers in terms of temperature, ionic strength, and solvent effects.Consider the titration of 50.0 mL of 0.170 M HA (K₂ = 4.2 x 105, pKa = 4.38) with 0.150 M NaOH. Determine the pH at the half-equivalence point 4.38 ○ 2.45 4.89 ○ 7.00 ○ 6.57