Why is KHP (MW = 204.22 g/mol) preferred over benzoic acid (MW = 122.12 g/mol) as a primary standard for the aqueous NAOH titrant? KHP is more stable in air compared to benzoic acid. O KHP has a larger molecular weight, hence there is smaller relative mass measurement error in weighing KHP than benzoic acid. KHP is less soluble in water compared with benzoic acid. O All of the above.
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- A 1.0-LL buffer solution initially contains 0.30 molmol of NH3NH3 (Kb=1.76×10−5)(Kb=1.76×10−5) and 0.30 molmol of NH4ClNH4Cl. What mass of the correct reagent should you add? Express your answer using two significant figures.A 0.15mL of phenolphthalein indicator was dropped to 10.00mL of an unknown acid. Then, 52.00mL from the 100mL of 10M NaOH Solution was placed in a 50mL burette. A titration experiment was done using the 10.15mL unknown acid with phenolphthalein indicator as analyte and the 52mL of NaOH as titrant. The solution turned pink when the final volume of the analyte was 10.68mL and the final reading in the burette was 51.47mL. What is the concentration of the unknown acid?In the titration of 25.00 mL of a water sample, it took 19.040 mL of 4.965x 10−3 M EDTA solution to reach the endpoint. The total hardness is always listed in parts-per-million (ppm) of CaCO3 (or mg CaCO3 / Kg H2O). Since the density of water is 1.0 g/mL, one ppm would be the same as the number of mg of CaCO3 per liter of water. Determine the number of moles of CaCO3 present in the titrated sample of water, assuming that all the Ca2+ combines with CO32−. (enter your answer with 3 significant figures)
- A student titrates a saturated solution of Ca(OH)₂ with 0.0100 M HCl. A 10.00 mL sample of the Ca(OH)2 solution required 20.04 mL of HCI to reach the endpoint of the titration. If there are 0.2004 mmoles of HCI added in the titration and 0.2004 mmoles of OH- present in the saturated solution, calculate [OH-] and [Ca²+] in the saturated solution.1 ) The density of a 5.26MNaHCO 3 (84.0 g/mol) is 1.19g / m * l . Its molality is 2) Calculate the pAg^ + at the equivalence point in the titration of 25.0ml of 0.0823 M Kl with 0.051M AgNO 3 . Ksp Agl=8.3*10^ -16 3) Commercial concentrated aqueous nitric acid is 70.4% HNO3(63.0 g/mol) by mass and has a density of 1.41g / m * l . The molarity of this solution is 4) Consider the titration of 25ml of 0.0823M KI with 0.051M AGNO3, Kspagi =8.3x10-16 Calculate pAg* after adding 39.0 ml I03 Ag* + 103 ====AglO3 5) Commercial concentrated aqueous nitric acid is 70.4 1\%HNO 3 (63.0 g/mol) by mass and has a density of 1.41g / m * l . The molarity of this solution is: 6) What mass in g of Na 2 CO 3 [106 g/mol] is required to prepare 250 ml of 0.3M aqueous solution in Na^ + [23.g/mol] ? 7) Calculate pAg^ + after adding 42.30ml AgNO 3 In the titration of of 0.0823M Kl with 0.051M AgNO 3 . KspAgl = 8.3 * 10 ^ - 16 8) The milliliters of concentrated HClO 4 (100.5 g/mol),6 60% by mass,…A 10.0 mL aliquot of H2 SO4 solution was diluted to 100.0 mL. Then 25.0 mL aliquot of the diluted H2 SO4 solution was titrated with 30.0 mL of 0.200 M NaOH solution using phenolphthalein as indicator. What is the molar concentration of the original H2 SO4 solution?
- A chemist is titrating a 32.5 mL sample of 2.00x10^-3 M benzoic acid with 8.00x10^-4 M Ca(OH)2. At what volume of Ca(OH)2 will the chemist reachthe analyte's equivalence point? a) 105 mL Ca(OH)2 b) 64.5 mL Ca(OH)2 c) 81.3 mL Ca(OH)2 d) 40.6 mL Ca(OH)2 e) 163 mL Ca(OH)2Calculate the mass of acetic acid in 100ml of both samples in vinegar. given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of H2SO4Volume of NaOH = 10.0 mL…What is the equivalence volume in the titration of 50.00 mL of 0.010 0 M NaOH with 0.100 M HCl? Calculate the pH at the following points: Va 0.00, 1.00, 2.00, 3.00, 4.00, 4.50, 4.90, 4.99, 5.00, 5.01, 5.10, 5.50, 6.00, 8.00, and 10.00 mL. Make a graph of pH versus Va.
- Calculate pAg at each of the following points in the titration of 20.00 mL of 0.0500 M AgNO3 with 0.0250 M NH4SCN.Ksp AgSCN = 1.10 × 10-12 a. 0.00 mL NH4SCNb. 20.00 mL NH4SCNc. 40.00 mL NH4SCNd. 45.00 mL NH4SCNe. 50.00 mL NH4SCNUse the titration data below to answer the following questions regarding the titration of an unknown acid with sodium hydroxide. a.)How many acidic protons are present in this unknown acid? b.For the volume change between 8 and 9 mL, what is the value of the first derivative (dpH/dV)? c.) What is the pKa1 for this unknown acid?Determining the ksp of Calcium Hydrioxides Titration of Calcium Hydroxide with HCl Mass of Erlenmeyer flask: 25.52 g Mass of Erlenmeyer Flask + Calcium Hydroxide Solution (lime water) (g): 28.44 g Mass of Calcium Hydroxide Solution (g): 2.92 g Volume of Ca(OH)2 Density = 1.000 g/mL (mL) - 2.92 mL Concentration of HCl (M) - How do you find this? HCl 0.10 M solution used Initial HCl volume = 0.79 mL Final HCl volume = 0.09 mL volume of HCl delivered = 0.70 mL Moles of HCl delieverd - how do I calculate this? Moles of OH- in sample - how do I calculate this? Moles of Ca2+ in sample - how do I calculate this? Thank you!