2. You need to make a solution that has pH of 4.6 from sodium acetate/acetic aci

Question

2. You need to make a solution that has pH of 4.6 from sodium acetate/acetic acid Calculate the needed concentrations if the sum of the molarities of sodium acetate/ac etic acid is 0.05 M. he next three questions pertain to the preparation of a buffer starting with the salt of the weak base and adding an amount of strong acid to convert some salt to its weak acid. The amount of strong acid is selected to achieve the desired amount of the conjugate pair concentration ratio at the specified pH 3. You need to prepare 100 mL of a solution that is 0.05 M sodium bicarbonate. Calculate amount sodium bicarbonate you need to weigh out to prepare this solution. You want to convert a portion of the bicarbonate in this solution to carbonic acid with a molarity of 0.02M utilizing 1 M HCl. How much HCI do you add? (Ignore the slight change in volume.) 4. 5. What will be the molarities of the two components in the solution? 6. Write three acid dissociation chemical equations for phosphoric acicd. 7. If you are preparing a phosphate buffer with a pH of 12.2, which chemical equation are you working with and what are the two components of the buffer?

Explanation / Answer

2. Prepare NaC2H3O2/HC2H3O2 buffer

pKa = 4.75

[HC2H3O2] + [NaC2H3O2] = 0.05 M

using hendersen-hasselbalck equation,

pH = pKa + log([NaC2H3O2]/[HC2H3O2])

4.6 = 4.75 + log([NaC2H3O2]/[HC2H3O2])

[NaC2H3O2] = 0.71[HC2H3O2]

or,

[HC2H3O2] + 0.71[HC2H3O2] = 0.05 M

[HC2H3O2] needed = 0.05/1.71 = 0.03 M

[NaC2H3O2] needed = 0.05 - 0.03 = 0.02 M

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3. amount of sodium bicarbonate = 0.05 M x 0.1 L x 84.007 g/mol

                                                     = 0.420 g

4. amount of 1 M HCl needed = 0.02 M x 100 ml/1 M = 20 ml

5. molarity of two component in he solution

[NaHCO3] = 0.03 M x 100 ml/120 ml = 0.025 M

[H2CO3] = 0.02 M x 100 ml/120 ml = 0.017 M

6. Dissociation equation for H3PO4

H3PO4 + H2O <==> H3O+ + H2PO4- ....... Ka1

H2PO4- + H2O <==> H3O+ + HPO4^2- ....... Ka2

HPO4^2- + H2O <==> H3O+ + PO4^3- ....... Ka3

7. for pH 12.2,

components : HPO4^2- and PO4^3-

equation,

HPO4^2- + H2O <==> H3O+ + PO4^3- ....... Ka3

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