Acids/Bases Buffer Practice CHEM 314 Acids/Bases, pH, and Buffers PRACTICE Sprin

Question

Acids/Bases Buffer Practice

CHEM 314 Acids/Bases, pH, and Buffers PRACTICE Spring 2018 1. Provide the overall charge and the structure (showing correct protonation state) for phosphoric acid, H3POs at: a. pH 5 b. pH 8 C. pH 10 d. pH 13 e. pH 1 2. Calculate the pH and percent ionization for a 0.500 M Formic Acid solution. See Table 2.4 in your text for the appropriate Ka value. ANSWER (pH): Answer(% ionization); 3. Prepared lactate buffer has significant concentrations of both lactic acid and its conjugate base lactate present. Determine the relative concentrations (ratio) of these two species in a solution having a pH of 4.20. Wold this buffer respond better buffer more effectively) to added acid or added base? Briefly explain using proper chemistry Ratio: Explain: 4. Acetic acid has a pKa of 4.8. How many milliliters of 0.2 M acetic acid and 0.2 M sodium acetate are required to prepare 1 liter of 0.1 M buffer solution having a pH af 4.8? Answer: Page 1 of2

Explanation / Answer

1. Overall charge anf structure at pH values for H3PO4 are shown below,

a. pH 5

Structure : H2PO4-

charge : -1

b. pH 8

Structure : HPO4^2-

charge : -2

c. pH 10

Structure : HPO4^2-

charge : -2

d. pH 13

Structure : PO4^3-

charge : -3

e. pH 1

Structure : H3PO4

charge : 0

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2. HCOOH = 0.5 M

HCOOH <==> H+ + HCOO-

let x amount dissociated

Ka = [HCOO-][H+]/[HCOOH]

1.8 x 10^-4 = x^2/0.5

x = [H+] = 9.5 x 10^-3 M

pH = -log[H+] = 2.02

percent ionization = 9.5 x 10^-3 x 100/0.5 = 1.9%

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3. lactate buffer

pKa = 3.86

ratio lactate/lactic acid

Hendersen-Hasselbalck equation,

pH = pKa + log(lactate/lactic acid)

4.20 = 3.86 + log(lactate/lactic acid)

ratio(lactate/lactic acid) = 2.2

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4. pH needed = 4.8

pKa acetic acid = 4.8

since pH = pKa

To prepare 1 L of 0.1 M acetate buffer,

equal portions of 0.2 M sodium acetate and 0.2 M acetic acid are to be mixed

total moles = 0.1 M x 1 L = 0.1 moles

so 0.05 mole acetic acid and 0.05 mole sodium acetate is needed

volume of acetic acid needed = 0.1 M x 1 L/2 x 0.2 M = 0.25 L

volume of sodium acetate needed = 0.1 M x 1 L/2 x 0.2 M = 0.25 L

Dilute the resulting solution to 1 L to get the final buffer solution

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