You are given a water sample and asked to determine the alkalinity for the sampl

ID: 906662 • Letter: Y

Question

You are given a water sample and asked to determine the alkalinity for the sample. The water is at pH 9 (before you begin titrating with 0.1 N HCl).

You perform 3 separate titrations each on a 100 mL aliquot of the sample.

Titration #1: Requires 6 mL to reach the H2CO3 (f=0) equivalence point.

Titration #2: Requires 0.3 mL to reach the NaHCO3 (f=1) equivalence point.

Titration #3: Requires 0 mL to reach the Na2CO3 (f=2) equivalence point.

a) Calculate the total alkalinity, phenolphthalein alkalinity, and caustic alkalinity for this sample.

b) Calculate the carbonate alkalinity and bicarbonate alkalinity.

c) Calculate the total carbon (C-total), [H2CO3 * ], [HCO3 - ], and [CO3 2- ] present in the water sample

pH = 9 = -log[H+]

[H+] = 1 x 10^-9 M

[OH-] = 1 x 10^-14/1 x 10^-9 = 1 x 10^-5 M

moles of HCl used for Na2CO3 = 0 ml

moles of HCl used for NaHCO3 = 0.1 x 0.0003 = 3 x 10^-5 mols

concentration of CO3^2- = 3 x 10^-5/0.1 = 3 x 10^-4 M

moles of HCl used for H2CO3 = 0.1 x 0.006 = 6 x 10^-4 mols

concentration of HCO3- = 6 x 10^-4/0.1 = 6 x 10^-3 M

K1 = 2 x 10^-4 = 1 x 10^-9 x 6 x 10^-3/[H2CO3]

[H2CO3] = 3 x 10^-8 M

a) Total alkalinity = [(HCO3-) + 2(CO3^2-) + [OH-]]/2

= (6 x 10^-3 + 2 x 3 x 10^-4 + 1 x 10^-5)/2 = 3.305 x 10^-3 mol/L

Caustic alkalinity = 50000 x (10^(-14+pH)) = 0.5 mg/L

b) carbonate alkalinity = 1 x 10^5 x (K2(HCO3-)(10^pH))

= 1 x 10^5 [5 x 10^-11 x 6 x 10^-3 x 10^9)]

= 30 mg/L

bicarbonate alkalinity = 5 x 10^4 x [2TA - 10(-14+pH))/1+2K2(10^pH)]

= 5 x 10^5 [2 x 3.305 x 10^3 - 10^-5/1 + 2 x 5 x 10^-11 x 10^9)]

= 30.091 mg/L

c) Total carbon = 3 x 10^-4 + 6 x 10^-3 + 3 x 10^-8 = 6.3 x 10^-3 mol/L

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