Hydrothermal vents in the ocean offer a unique environment for growth in which o
ID: 1049699 • Letter: H
Question
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. 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_2 COONa, molar mass 116.48 g mol^1) 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) 2.00 L of a starting solution. 1.25 mol L^-1 acetic acid (CH_3 COOH, pK_a = 4.74) and solid sodium acetate (CH_3 COONa, molar mass 82.04 g mol^-1) Which buffer system would best suit the needs of the scientists studying hydrothermal vents? Clearly indicate your choice with a tick. 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? pH = pKa + Log^[Salt]/[acid] 2.50 = 2.85 + log [SaH]/1.25 => Log [Salt]/1.25 = -0.30 => [Salt]/1.25 = 0.5012 => [Salt] = [conjugate base] = 0.6265 M What mass (grams) of the Solid (ie depending an 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
(c) In chloroacetic acid / sodium choloroacetate buffer, the molarity of the salt sodium chloroacetate = 0.6265 M
Total volume of the buffer = 2 L
Moles of salt in the buffer solution = ( Molarity x volume in L) = ( 0.6265 x 2) moles = 1.253 moles
Molar mass of sodium chloroacetate = 116.48 g/mol
Mass of 1.253 moles of sodium chloroacetate = ( 116.48 x 1.253) g = 145.94 g = 146 g
The required mass of soild sodium chloroacetate = 146 g
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