Three titration trials were performed and three molar concentrations of NaOH were calculated. The three molar concentrations are as follows: 0.10 M 0.12 M 0.19 M Calculate the standard deviation of the molar concentration of NaOH.
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- Trial 1: Initial pH= 3.92mL Mass of KHP and Paper= 0.868g Mass of paper= 0.357g Mass of KHP= 0.511g Trial 2: Initial pH= 4.09ml Mass of KHP and Paper= 0.870g Mass of paper= 0.359g Mass of KHP= 0.511g 1)Use Kb, the number of moles of C8H4O4^2- at the equivalence point, and the total volume at that point to calculate the pH for each sample at the equivalence point. Compare these calculated results with the experimental results.Please show work A tums tablet pulverized (powdered) and 0.4265 g of this power was mixed with 9.00 mL 1.000 M HCl. This mixture was back titrated with 0.1000 M NaOH with phenolphthalein as indicator. It required 8.25 mL of 1000 M NaOH to reach the end point. Compute % CaCO3 in the tums sample and % error.Table 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.19 2.26 2.39 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 48.55 49.43 49.99 Expected color at end point Volume of NaOH used (mL) Average Volume of NaOH used in liters Average moles of NaOH used (mol) Average moles of acetic acid (mol) Average molarity of acetic acid (M) Average mass of acetic acid (g) Average mass of vinegar (g) (assume the density of vinegar is 1.00 g/mL) Average mass % of acetic acid in vinegar Known mass % of acetic acid in vinegar is 5.45% Percent Error
- A 250.00 mL solution of 0.00215 M AB4 is added to a 230.00 mL solution of 0.00380 M C3D2. What is pQsp for A3D4? Answer: (18.607)Trial 1: Initial pH= 3.92mL Mass of KHP and Paper= 0.868g Mass of paper= 0.357g Mass of KHP= 0.511g Trial 2: Initial pH= 4.09ml Mass of KHP and Paper= 0.870g Mass of paper= 0.359g Mass of KHP= 0.511g 1) find the volume at equivalence point of both trial 2) find the pH at equivalence point of both trial and calculate the molarity of the NaOH solution from each result and calculate the mean.The water hardness of tap water has been determined to be 120 (= 120 mg CaCO3 per 1 L of water). After the tap water has run through the ion exchange column, titration of a 20.00 mL sample used only 0.70 mL of 0.0100 M EDTA, to reach the endpoint. Calculate the effectiveness of the ion exchange column: what is the % of the calcium, removed by the ion exchange column? Provide your answer in % and without decimal. ( Please type answer note write by hend)
- Review the list of common titration errors. Determine whether each error would cause the calculation for moles of analyte to be too high, too low, or have no effect. ANSWER BANK: Too low, No effect OR Too high adding titrant past the color change of the analyte solutionrecording the molarity of titrant as 0.1 M rather than its actual value of 0.01 Mspilling some analyte out of the flask during the titrationstarting the titration with air bubbles in the buretfilling the buret above the 0.0 mL volume markFast pls i will give u like for sure solve this question correctly in 5 min pls A student analyzed a CaCO3 antacid tablet by neutralizing the base with an excess of .60 M HCL and back-titrating the excess with .50M NaOH. He recorded the following data: mass of tablet = 1.3g amount of active ingredient CaCO3 per label = 500 mg volume of HCl added: 25 mL initial buret reading : 34 mL final buret reading : 47.4 mL If the concentration of the provided HCl is actually higher than .60M, would the determined mass of CaCO3 in the antacid tablet be too high or too low?Data Table 1: Complexometric Titration of Hard Water 6ml of water used (5ml of hard water and 1ml of buffer) Hard Water Trial 1 Hard Water Trial 2 Hard Water Trial 3 Initial Syringe Reading (mL) 1ml 1ml 1ml Final Syringe Reading (mL) 0.55ml .50ml .33ml Volume of EDTA Consumed (mL) 0.45ml .50ml .77ml Water Hardness ppm CaCO3 (mg/L) Average ppm CaCO3 (mg/L): photo shows equation that needs to be used to solve
- Data Table 3: Complexometric Titration of Soft Water 5 ml of water 1ml of buffer Hard Water Trial 1 Hard Water Trial 2 Hard Water Trial 3 Initial Syringe Reading (mL) 1ml 1ml 1ml Final Syringe Reading (mL) 77ml 79ml 75ml Volume of EDTA Consumed (mL) .23ml .21ml .25ml Water Hardness ppm CaCO3 (mg/L) Average ppm CaCO3 (mg/L):Based on the analysis of a vinegar solution via titration with 0.1 M NaOH. Results are as follows: Trial 1 Volume of vinegar solution used (mL) 5.00 Final burette reading (mL) 44.20 Initial burette reading (mL) 0.00 Determine the percentage mass of acetic acid in vinegar for Trial 1. The density of vinegar is 1.01 g/mL and molecular mass for acetic acid is 60 g/mol.A 50.00 (±0.02) mL portion of an HCl solution required 29.71(±0.02) mL of 0.01963(±0.0032) M Ba(OH)2 to reach an end point with bromocresol green indicator. ? of HCL = 29.71?? ? 0.01963 ???? ??(??)2 ?? ? 2 ???? ??? ???? ??(??)2/ 50.00?? = 0.02333 ? Calculate the uncertainty of the result (absolute error).Calculate the coefficient of variation for the result.