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Temperature Dependence of Ka for a Weak Monoprotic Acid Temperature Dependence o

ID: 822764 • Letter: T

Question

Temperature Dependence of Ka for a Weak Monoprotic Acid

Temperature Dependence of Ka for a Weak Monoprotic Acid Question: Calculate the grams of sodium chloroacetate. NaClCH2COO: needed to prepare 50 ml of a 0.256 M NaClCH2COO solution. What is a buffer? What is a buffer made up of? Give two (2) examples of buffer solutions. Define Le Chatelier's principle. Explain how temperature affects a system in equilibrium. The pKa of acetic acid at room temperature is 4.75. What is the pH of a solution of acetic acid at half the equivalence point? Calculate the grams of sodium chloroacetate, NaClCH2COO needed to prepare 25mL of a 0.198M NaClCH2COO solution.

Explanation / Answer


Moles of NaClCH2COO = volume x concentration = 50/1000 x 0.256 = 0.0128 mol

Mass of NaClCH2COO = moles x molar mass of NaClCH2COO

= 0.0128 x 116.48 = 1.49 g


(1) A buffer is a solution that is able to resist pH changes due to addition of small amounts of acid or base.


(2) A buffer is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid.


(3) Examples of buffer solutions are:

(a) acetic acid (CH3COOH)/sodium acetate (NaCH3COO)

(b) ammonia (NH3)/ammonium chloride (NH4Cl)


(4) Le Chatelier's principle states that if a system in equilibrium is subjec to a change (e.g in concentration), it will adjust itself to counteract that change by establishing a new equilibrium.


(5) Increasing temperature adds heat while decreasing temperature removes heat from the system and disrupts the equilibrium

=> system will shift to adjust to the addition or removal of heat (depending on whether reaction is endothermic or exothermic)


(6) At half equivalence point: [CH3COO-] = [CH3COOH] => [CH3COO-]/[CH3COOH] = 1

pH = pKa + log([CH3COO-]/[CH3COOH]) = pKa + log(1) = pKa = 4.75


(7) Moles of NaClCH2COO = volume x concentration = 25/1000 x 0.198 = 0.00495 mol

Mass of NaClCH2COO = moles x molar mass of NaClCH2COO

= 0.00495 x 116.48 = 0.577 g

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