Hello, I answered these quesitons but I want to chekc to see if I got the right

Question

Hello, I answered these quesitons but I want to chekc to see if I got the right answer and did it the correct way. Thank you for your time.

1.The alkalinity of natural waters is usually controlled by OH-, CO32-, and HCO3-, which may be present alone or in combination. Titrating a 100.0 mL sample to a pH of 8.3 requires 18.70 mL of a 0.0281 M solution of HCl. A second 100.0 mL aliquot requires 48.20 mL of the same titrant to reach a pH of 4.5. Calculate the concentrations of CO32- and HCO3- in ppm.

2.Confirm the math to confirm that 8.7 mL of 37.2% HCl diluted into 1L equals ~0.1 M HCl?

3. Wouldn't a graduated cylinder be the best glassware to transfer 25mL of aliquot from a vol. flask to a 125mL Erlenmeyer flask?

Explanation / Answer

Answer:

1) As mentioned there is no hydroxide ion concentration in the solution

the 100mL sample required 18.70mL of HCl solution to reach the pH =8.3

Hence, The first equivalence point is due to half neutralization of carbonate

CO3-2 + H+ ---> HCO3-

concentration of bicarbonate = concentration of acid used X volume of acid used / total volume

concentration of bicarbonate = 0.0281 x 18.70 / 100 = 0.0053 Moles / Litre

Second equivalence point is due to complete neutralization of HCO3- and bicarbonate, the pH = 4.5

CO3-2 + H+ ---> HCO3-

HCO3- + H+ --> H2O + CO2

so total concentration of carbonate and bicarbonate ions will be

volume of natural water X concentration = volume of HCl used X concentration of HCl

Concentration of natural water = 48.20 x 0.0281 / 100 = 0.0135

[CO3-2] = 0.0053 moles / Litres = 0.0053 X molecular weight (g / L) = 0.318 g/L = 318 mg / L = 318ppm

2X [CO3-2] + [HCO3-] = 0.0135

[HCO3-] = 0.0029 moles / Litres = 0.0029 X 61 g /L = 0.177 g /L = 177 mg / L = 177ppm

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