C- Al is determined by titrating with EDTA Al +H;Y AIY +2H 0.5 mg sample required 20.5 ml. EDTA for titration. The EDTA was standardized by titrating 25 ml. of 0.01M CaCl; solution, requiring 30 ml. EDTA. Calculate AlO, in the sample.
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- Plot the titration curve for titration of 50mL of 0.0150M Fe2+ solution buffered to pH's 7.0 with 0.03M EDTA. A-) OmL, B-) 10mL, C-) 25 mL, D-) Calculate the pFe after 26 mL of titrant additions. (Kr.Y= 2.1x10^14, a4=4.8x10^-4)A 100 mL sample of drinking water was buffered at pH 10 and after a ddition of Eriochrome Black T required 1 1 mL of 0.00499 M EDTA for titration. Identify the color at the end point.2- Ahmed has done four types of titrations of 1M and 50 ml for both acid and base (Conductometric Titration of Strong Acid with Strong Base , Potentiometric Titration of Weak Acid with Strong Base , Potentiometric Titration of Strong Acid with Strong Base and Conductometric Titration of Weak Acid with Strong Base ). He plotted the results in curves as shown below. Match between the type of titration and his results 3- Ahmed standardized NaOH (10 ml) with KHP( mass=1 g , MW= 204 g/mol ). Calculate the concentration of standardized NaOH? 4- Calculate pka for a weak acid titrated with strong base using the curve shown below? 14 12- 10- 8 6- 4- 2- 0- 10 20 30 40 50 5- If the solute moved 3 cm in paper chromatography and R, was 0.5. How long in cm the solvent has moved?
- concentration of I3- : 0.002M d. Titration data: Final volume: 29.05 mL Starting volume: 38.10 mL Delivered volume: 25.15 mL How do you calculate g?4. A standard solution of EDTA (0.100 M) is being used to titrate 25.00 ml of a 0.10 M Zn solution (buffered at pH 8.0). Calculate the value of pZn": a) before the addition of EDTA solution b) after the addition of 10.00 ml of EDTA solution c) at volume V = } Veq (Veg = volume at equivalence point) d) at volume = Veg (at the equivalence point) e) at volume V = 2,0 VegO Assignm Assigned Jessica Marasigan Example: 100 points d or create A 0.2420 g sample cntg. Calcium is dissolved and the metal precipitated as CaC,O4 The ppt. is filtered, washed and redissolved in acid. The pH is adjusted, 25 ml Of 0.0400 M rk as done EDTA added and the xcss EDTA titrated with 33.28 ml of nments 0.01202 M Mg²*. Calc. the % Ca (40.078) in the sample. O Class comme o Jessica Marasigan Add a class comm
- A 25.00 mL unknown water sample is titrated with a standardized 0.0140 M EDTA solution. It is determined that 8.54 mL of EDTA is required to reach the end point. The blank titre was determined to be 1.50 mL. Determine the concentration of Calcium in the water sample in ppm. (mwt. of CaCO3 = 100.0892 g/mol).The concentration of CA2+ in an unknown solution was determined by titration with the EDTA solution standardized with a molarity of 0.000085 M. 25 mL of the unknown solution was diluted to 250 mL with deionized water. A 25 mL aliquot of the diluted solution was then titrated with the EDTA solution. The titration required 35.54 mL of the EDTA solution to each the end point. What is the molarity of calcium in the original unknown solution?PERIMENTAL ANALYTICAL CHEMISTRY weill & ne My courses EXPERIMENTAL ANALYTICAL CHEMISTRY General Quiz2 Time Upon titration of 5.0 ml solution of Zn2*15.0 ml of 0.05M EDTA was required, then the concentration of Zn2* is 0.15 M. ut of
- A 0.4071-g sample of CaCO3 (MM: 100.09 g/mole) is transferred to a 500-mL volumetric flask, dissolved using a minimum of 6 M HCl, and diluted to volume. A 50.00-mL aliqout of this solution was mixed with 5 mL of a pH 10 NH3-NH4Cl buffer that contains a small amount of Mg2+--EDTA to adjust the pH to 10. After adding calmagite as an indicator, the solution is titrated with the EDTA, requiring 42.63 mL to reach the end point. Report the molar concentration of EDTA in the titrant. 9.543 x 10-3 M 5.748 x 10-3 M 3.266 x 10-4 M 1.018 x 10-4 MCalculate pre at cach of the points in the titration of 25.00 mL of 0.02206 M Fe by 0.03501 M EDTA at a pH of 5.00, The valucs for log K, and ay can be found in the chcmpendix. 10.00 mL pFe the equivalence point, V. 19.00 ml. E TOOLS X10A 0.4755-g sample containing (NH,),C,O, and inert materials was dissolved in water and 3. strongly alkaline with KOH, which converted NH,* to NH3. The liberated NH; was distilled into 50.0mL of 0.05035 M H,SO The excess H,SO, was back titrated with 11.13 mL of 0.1214 M NaOH. Calculate the following: A. The % N (14.00 g/mole) B. The % (NH,),C,0;(124.10 g/mole) in the sample
