Please Answer ALL. thanks. Hydrothermal vents in the ocean offer a unique enviro

Question

Please Answer ALL.

thanks.

Hydrothermal vents in the ocean offer a unique environment for growth in which only specialist organisms, such as the Pompeii worm, may grow due to the extreme temperature, pressure and pH conditions as well as the range of toxic chemicals found. To simulate the conditions near a particular ocean vent, an aquatic chemist wishes to make a solution in which the pH will be held constant at 2.50. The following materials are available. I. 2.00 L of a starting solution, 1.25 mol L^-1 chloroacetic acid (ClCH_2COOH, pK_a = 2.85) and solid sodium chloroacetate (ClCH_2COONa, molar mass 116.48 g mol^-1) II. 2.00 L of a starting solution, 1.25 mol L^-1 nitrous acid (HNO_2, pK_a = 3.15) and solid sodium nitrite (NaNO_2, molar mass 69.00 g mol^-1) III. 2.00 L of a starting solution, 1.25 mol L^-1 acetic acid (CH_3COOH, pK_a = 4.74) and solid sodium acetate (CH_3COONa, molar mass 82.04 g mol^-1) a) Which buffer system would best suit the needs of the scientists studying hydrothermal vents? Clearly indicate your choice with a tick. b) What concentration of the conjugate base (ie depending on your answer to part a: either sodium chloroacetate, sodium nitrite or sodium acetate) is required to produce an appropriate buffer using the starting solution given?

Explanation / Answer

1) the required buffer pH = 2.50

so we will use the buffer system with nearest value of pKa to the desired pH

Hence we should use chloroacetic acid and sodium chloroacetate as ideal buffer with pKa = 2.85

2) Now we will use Hendersen's Hassalbalch's equation to determine the the mass of conjugate base

pH = pKa + log [salt] / [acid]

pKa = 2.85

pH = 2.5

Moles of acid = Molarity X volume = 1.25 X 2 = 2.5 moles

Moles of salt = ?

2.5 = 2.85 + log [salt] / [2.5]

-0.35 = log [salt] / [2.5]

Taking antilog

0.447 = [salt] / [2.5]

Moles of salt required = 1.118

Mass of salt required = Moles X molecular weight = 1.118 X 116.48 g / mole = 130.22 grams

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