What is the pH of a 0.0100 M solution of analine (C6H5NH2) , a weak base. Kb= 4.0 x10-10 (ICE problem) Your answer should have 2 digits after the decimal
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What is the pH of a 0.0100 M solution of analine (C6H5NH2) , a weak base.
Kb= 4.0 x10-10 (ICE problem)
Your answer should have 2 digits after the decimal
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- A 0.345 M of generic weak acid was prepared and has a Kb of 3.91 x 10-10. What is the pH of this weak acid? Use ICE tableProblem Solving. Solve the following problems, use GRESA format in answering A 0.0001 molar basic solution is 1.50 % ionized. What is the pH of the solution if its Kb= 1.25 × 10^−6?a MONOPROTIC ACID: benzoic acid (C6H5CO2H) ka: 6.5 x 10^-5 find for the pH of a solution that 0.50 M. Show all work please. (Use and ICE table if able too)
- Parameter's Starting HCl conc. (M) 1.000 mL of HCl added 25.00 mL of NaOH to endpoint 21.50 Starting NaOH conc. (M) 1.000 mL of Base Added pH Observed pH Calculated mmol of HCl mmol NaOH mmol of Excess Total Volume M(H+) M(OH-) pOH 18.00 1.30 19.00 1.50 20.00 1.75 20.10 1.80 20.20 1.89 20.30 1.95 20.40 2.01 20.50 2.05 20.60 2.07 20.70 2.09 20.80 2.11 20.90 2.20 21.00 2.27 21.10 2.54 21.20 2.76 21.30 3.30 21.40 5.30 21.50 7.00 21.60 8.40 21.70 10.12 21.80 11.40 21.90 12.10 22.00 12.45 23.00 13.25 24.00 13.54 25.00 13.71 26.00 13.85 27.00 13.93 please help calculate for the rest!PS. Further values required for the solvings are give in the various situations below. (ANSWER) Situation: A community in a mountainous area of Bohol uses water collected from a nearby natural spring. A sample was submitted to a laboratory for the analysis of its total hardness. Required: SHOW YOUR COMPLETE CALCULATIONS. BASED ON THE IMAGE PROVIDED BELOW FOR THIS QUESTION: Calculate the amount of titrant used in each trial to reach endpoint. Report total hardness of the sample as mean ±sd. a. 250.0 mL of 500.0 ppm of CaCO3 solution from a primary standard (assume solvent is distilled water only). Answer : Mass of CaCO3 = 0.125 g b, The EDTA solution was standardized by titrating it with a 25.0 mL aliquot of the CaCO3 solution. How much of the titrant was consumed. Answer: Volume of EDTA consumed = 12.405 g c. Calculate the average titer (mg CaCO3/mL EDTA). Mass of CaCO3 in 25 mL CaCO3 solution: 0.0125 g Answer: 1.008 mg CaCO3/ mL EDTATopic: Standardization of acid and base with back titration Note: Include up to 4 decimal places Kindly explain the process Thank you!
- Why is the color violet/blue at pH near 7? Answer in 1 sentence.What is the percentage of total acid expressed as acetic acid (CH3COOH) in a sample of vinegar if 3.000 g of the vinegar requires 20.50 mL of 0.1150 N KOH solution for an endpoint with phenolphthalein indicator? (Note: Use the least significant digits in your answer.)please give calculation as well. Calculate the pH value of each of the solutions in tubes 1-9 using the Henderson-Hasselbalch equation (H-H eqn).Determine the pl of casein. Compare your experimental values with those found in the literature. Biomolecules. Thank you! The protein is casein.
- What is the pH of the solution after mixing 0.171 g of Mg(OH)2 (MW=58.321 g/mol) with 68.9 mL of 0.0569 M HCl? The resulting solution was diluted to 100 mL. Round your calculated value for pH to two figures to the right of the decimal point.T or F: The normality of the solution obtained by dissolving the acid sample be approximately the same as that of the titrant.pka=pH halfway to the equivalence point. what is the equivalence point volume? it says to do so by locating the inflection point of the curve. but i'm not sure what the inflection point of the curve is. it says: using a ruler, draw a vertical line that crosses the inflection point and extend this line so it crosses the horizontal axis (use vertical minor gridlines to guide vertical ruler) to find volume base at the equivalence point.