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Hardness water lab help please fill in the blanks thank you. will thumb up answe

ID: 495646 • Letter: H

Question

Hardness water lab help

please fill in the blanks thank you. will thumb up answers.

Part A: Preparation of Calcium Carbonate Standard a) Mass of calcium carbonate b) Total volume of standard solution c) Molarity of calcium carbonate standard Part B: Determination of Blank Volume Trial 1 Trial 2 Trial 3 1aas Final Burette Reading (mL) Initial Burette Reading (mL) 5. 0,30 30 Volume for Blank (mL) Average Blank: LOO t Part C: Standardization of EDTA Titration Trial 1 Trial 2 Trial 3 Final Burette Reading (mL) Initial Burette Reading (mL) o 00 13 50 24ioo Volume for EDTA (mL) LOO Average Volume of Blank (mL) Volume of EDTA used to titrate Ca lilo ,8 I, 88 Molarity of EDTA (show calcs) Awrage Molarity of EDTA (show calc): 134 SM t eia3 (M)

Explanation / Answer

Moles of EDTA required to titrate 25mL unknown sample :

Trial 1 : 1.845 M 0.005 L= 9.23*10^-3 moles

trial 2 : 1.845 M * 0.0052 L= 9.59*10^-3 moles

Trial 3 : 1.845 M * 0.0053 L = 9.78*106-3 moles

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moles of caCO3 / 25mL of unknown solution

Trial 1 : 9.23*10^-3 moles

trial 2 : 9.59*10^-3 moles

Trial 3 : 9.78*106-3 moles

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weight of caCO3 per 25mL solution

Trial 1 : 100g/mol * 9.23*10^-3 moles = 0.923 gm

trial 2 : 100g/mol * 9.59*10^-3 moles = 0.959 gm

Trial 3 : 100g/mol * 9.78*106-3 moles = 0.978 gm

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weight od CaCO3 in 10^6 mL solution

Trial 1 : 0.923 gm *10^6/25mL = 36.92 *10^3 gm

trial 2 : 0.959 gm *10^6/25mL = 38.36*10^3 gm

Trial 3 : 0.978 gm * *10^6/25mL = 39.12*10^3 gm

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1 ppm = 1mg/L

trial 1 : 36.92 *10^3 gm / 10^6 mL

          = 36.92 *10^3 ppm

Trial 2 : 38.36 *10^3 ppm

Trial 3 : 39.12 *10^3 ppm

         

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