Question

please show all calculations and formulas

1. The following calibration plot was generated for you: Calibration of Calcium Concentration by AA Absorbance Value at 422.7 nm y = 0.00937x + 0.01323 RP = 0.99844 10 50 60 20 30 40 Calcium Concentration (ppm) Prior to running your sample on the AA instrument, you diluted your sample 1:1 with distilled water. You measured the calcium absorbance to be 0.2955. a) What is the concentration of calcium (in ppm) of your original water sample (not your diluted water sample)? b) Convert this calcium concentration into its equivalent concentration of CaCO3 (in ppm CaCO3). Show all steps. [The molar mass of CaCO3 is 100.09 g/mol and the molar mass of Ca2+ is 40.08 g/mol.) c) Given that your magnesium concentration contributed 115.81 ppm CaCO3, what is the total hardness of your water sample (in ppm CaCO)? add

Answer 1

Solution! (a) Calculation Cat2 in ppm from given absorbace 0.2955 B as follows: lostoral from the graph y za 10.00937) + 0.01323 y is the where (for unknown cubsorbance sample) and oc thue concentratim in unknown sample OC = -0.01323 0.00937 0 = 31.3 ppm Cath Here we the solution of unknownd sample is taking half and add half diluent, so our concentration B double | 31-3 x 2 = Co. ca . 62.6 ppm in unknow sample. (b) Conversion of Cate concenbution in terms of Calog og Follows: 40.oong/L cat contain & 100.09 mg/l Caco : 62-6 mg/l Cat? contain = 62.6x100.09 40.08 6.32 mg/h or ppm] CaCO3 (6 Total hardness of sample B as follows Given Emg+2] = 115.84 mgie - I cath] = 62.6 mg/L Using [caco J = 2.5 [cat] + 4.1 [mg + 2] = 12.5 62.6) +(4.1 x 115.81) Total Hardness = 633.2 mgn of cacaz)