Calculate the pCa for the titration of 50 ml of 0.02 M Ca2+ at pH = 8 with 0.4 M EDTA at the equivalence point Ca2+ + Y4- ⇄ CaY2- Kf = 5x1010 Select one: a. 7 b. 12 c. 10 d. 5
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Calculate the pCa for the titration of 50 ml of 0.02 M Ca2+ at pH = 8 with 0.4 M EDTA at the equivalence point
Ca2+ + Y4- ⇄ CaY2- Kf = 5x1010
Select one:
a. 7
b. 12
c. 10
d. 5
e. 9
Step by step
Solved in 2 steps with 2 images
- On titrating 20 ml of 0.1 M KCl with 0.1 M AgNO3. (For AgCI, Ksp = 1.82 X 10- 10). 1.The volume of AgNO3 titrant need to reach the equivalence point is ………...............ml. 2.At the equivalence point the value of pAg = ……………………. 3.pAg after addition of 25 ml AgNO3= …………………………………The amount of calcium in physiologic fluids can be determined by a complexometric titration with EDTA. In one such analysis, a 0.100-mL sample of blood serum was made basic by adding 2 drops of NaOH and titrated with 0.00528 M EDTA, requiring 0.233 mL to reach the end point. Report the concentration of calcium in the sample as miligrams of Ca per 100 mL. (Ca = 40.078 amu).A RbOH solution is titrated four (4) times against potassium hydrogen phthalate (KHP; FW=204.224) samples to the Phenolphthalein endpoint. Using the data below, determine the concentration of the RbOH solution? g of KHP Volume of Base Required 0.5373 g 42.49 mL 0.5856 g 43.88 mL 0.5790 g 48.56 mL 0.5856 g 44.60 mL (Report your answer as "mean +/- std dev") M What is the percent relative standard deviation? % What is the 99% Confidence Interval for the concentration of the solution (population mean)?
- A 50.00-mL aliquot of a 0.100 0 M I− solution was titrated with 0.100 0 M AgNO3. Calculate pAg+ when 45.00 mL 0.100 0 M AgNO3 was added. The Ksp of AgI is 8.3 × 10−17. a) 14.58 b)9.82 c)12.10 d)14.08 e)13.801. The molar solubility of Ag2SO3 is 1.55× 10-4 M. Calculate the Ksp of Ag2SO3. a. 7.45 x 10^-12 b. 1.55 x 10^-4 c. 2.40 x 10^-8 d. 1.49 x 10^-11 e. 3.72 x 10^-12 2. What is the difference between the end point of an acid–base titration and the equivalence point? a. There is no difference; these are just two terms meaning the same thing. b. At the end point, the acid and base are present in equal numbers of moles. At the equivalence point, the indicator changes color. c. The end point is always at pH 7 but the equivalence point can be at various pH values. d.The end point can be at various pH values but the equivalence is point always at pH 7. e. At the end point, the indicator changes color. At the equivalence point, the acid and base are present in equal numbers of moles A solution at 25 °C that contains 1× 10-8 M H3O+ is a. basic b. acidic c. neutral d. both acidic and basicChemistry written by hand. In titration of 50 ml 0.1 M Mn2+ with 0.01 M EDTA (pH = 7), what is pMn2+ when 100 ml EDTA is added? a.7 b.11 c.9 d.15 e.8
- Maria wants to determine the concentration of Calcium ions in water from their well. She used a 100.0 mL water sample, and its pH was adjusted to 10 and 3 drops of the indicator were added. She used 2.500 mL of 0.09945 M EDTA to reach the endpoint. What is the concentration (mg/L) of calcium ion in the sample? Identify the classification of water given the reference. Reference: 0 to 60 mg/L Ca is classified as soft, 61 to 120 mg/L as moderately hard; 121 to 180 mg/L as hard; and more than 180 mg/L as very hardA standard serum sample containing 102 meq/L chloride was analyzed by coulometric titration with silver ion. Duplicate results of 101 and 98 meq/L were obtained. (a) Calculate the mean error (b) Calculate the relative error (c) Calculate the relative accuracy100.0 mL sample of drinking water was buffered with ammonia at pH 10 and after addition of EBT required 21.4 mL of 0.00512 M EDTA for titration. FW CaCO3 = 100.09 g/mol. What will be the change in color at the end point? What is the water hardness in terms of ppm CaCO3?. What is the role of ammonia buffer in the analysis?
- A student titrated a 25.00 mL sample of water with 0.0200 M EDTA. The titration required 20.40 mL of EDTA. Calculate the ppm hardness in the water sampleYou want to measure the concentration of carbonate (CO32-) in a mildly basic solution by using an EDTA back titration. CaCO3 has a Ksp of 5x10-9. You add 50.00 mL of 0.3484 M CaCl2 to 500.0ml of sample and filter the solution to remove the precipitate. You then take 250.0 mL of the filtered solution and titrate with 0.1786 M EDTA. You require 23.72 mL to reach the endpoint. What is the concentration of carbonate in the original sample?A solution containing 60 mL of a 0.025 mol / L metal ion (Mn +) buffer buffered to pH 7.0 was titrated with a 0.05 mol / L EDTA solution. Data: Conditional formation constant (Kf)=10^15. Determine: i) The concentration of the free metal when ½ of the equivalence volume is added. ii) the concentration of free metal in the equivalence volume. iii) the concentration of free metal with an excess of 2 mL of EDTA? (Data - EDTA constants: K1=0.01; K2=2.19.10^-3; K3=6.92.10^-7; K4=5.75.10^-11)