The cyanide ion (CN) is highly toxic, but can be remediated in solution by redox reaction with the hypochlorite ion (the principal component of bleach). The balanced net ionic equation for this process is: 2 CN + 5 CHO' + H2O -->5 Cl + N2 + 2 HCO
A 3.5 x 10' L sample of water was found to be contaminated with cyanide ions at a concentration of 25uM.
A. How many grams of cyanide were dissolved in the total sample?
B. Assuming that the standard concentration of bleach for use in wastewater remediation is 0.20M, what volume was necessary to consume all of the cyanide in the sample?
C. Would the remediation of the sample above likely be conducted using the exact quantity of bleach solution calculated in your answer to the above question? Explain.

1. 2.28 * 10^3 g of CN-

2. 175 L of bleach

3. No. Because the bleach is consumed by ammonia produced in the reaction.

Explanation:

Concentration = mass / (molar mass * volume)

Mass = concentration * molar mass * volume (L)

Concentration = 25uM = 25 * 10^-6 M; molar mass of CN- = 26 g/mol; Volume = 3.5 * 10^6 L (Note: this is the exact volume of water in the full question)

Mass = 25 * 10^-6 * 26 g/mol * 3.5 * 10^6 L = 2.28 * 10^3 g of CN-

2. Equation of reaction

2CN-1 + 5ClO-1 + H2O ---> 5Cl-1 + N2 + 2HCO3 -1

Molar ratio of ClO- to CN- = 2:5 = 0.4

Concentration of ClO-, Ca = 0.20 M

Concentration of CN-,Cb = 0.000025 M

Volume of ClO-, VA = ?

Volume of CN- solution, Vb = 3.5 * 10^6 L

Using the formula; CaVa/CbVb = mole ratio

Va = CbVb * 0.4 / Ca

Va = (0.000025 * 3.5 * 10^6 * 0.4)/0.20

Va = 175 L

Therefore, volume of bleach required is 175 L

3. No. The consumption of hypochlorite bleach solution is far greater than that calculated from the balanced equation of reaction. This is because of the chlorination of ammonia that is formed during the cyanide detoxification.

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it's filtration, but this is an assumption so my apologies if i got this