29 - Midterm 3 Review-1 (4)

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Midterm 3 Review These are answers to the review problems You will also see a few “further questions” to think about designated with a * in red
1. What would the curve look like if a strong base was added to a buffer solution? At first the buffer solution would neutralize the base so the pH would stay constant. Once the buffer capacity was reached, then the base added would no longer be neutralized and the pH would increase (become more basic) Volume Base pH
2. What is the purpose of a buffer? Buffers resist change to pH so they can help keep a system within a small pH range
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3. A buffer contains 0.85M HC 8 H 7 O 2 (K a = 4.9 x 10 -5 ) and 0.46M NaC 8 H 7 O 2 . Calculate the pH of this buffer. ࠵?࠵? = ࠵? ࠵? ! + ࠵?࠵?࠵? A − HA ࠵? ࠵? ! = − ࠵?࠵?࠵? ࠵? ! Since it is a buffer system you can use H-H equation First identify the acid, HA, and base, A - ࠵?࠵? = ࠵?࠵? ( + ࠵?࠵?࠵? Base Acid ࠵? ࠵? ( = − log(4.9 x 10−5) = 4.3 ࠵?࠵? = 4.3 + ࠵?࠵?࠵? 0.46 0.85 ࠵?࠵? = 4.3 + ࠵?࠵?࠵? 0.54 ࠵?࠵? = 4.04
4. Sketch a titration curve for the following cases: A. A strong acid titrated into a strong base (scenario 2 below) Starting pH high (strong base), equivalence point at pH 7 B. A strong acid titrated into a weak base (scenario 4 below) Starting pH high, equivalence point <7 C. A strong base titrated into a weak acid (scenario 3 below) Starting pH low, equivalence point >7
5. Assuming each curve is for the same volume of analyte, what do you need to compare across titration curves to determine if they have the same or different: A. Initial concentration in the analyte? Compare the moles of titrant needed to reach the equivalence point (when it neutralizes all of the analyte). Moles = concentration x volume. If the concentration of titrant is the same for each you can compare just the volume needed to reach the equivalence point. B. Species in the analyte? Compare the pKa, which is the pH at the ½ equivalence point A. Degree of dissociation in the analyte? Compare the initial pH since that shows how much H + or OH - is released into solution. The more dissociation, the higher for base or lower for acid the pH will be.
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6. What is the purpose of an indicator in a titration? How would you select an indicator to use for a specific titration?
7. Calculate the volume of HCl needed to reach the equivalence point when 100 mL of 0.1M Ba(OH) 2 is titrated with 0.05M HCl. (Hint: think about what concentration of OH - ) C analyte V analyte = C titrant V titrant mol H + or OH - in sample mol H + or OH - in titrant (0.1M *2 )(0.1L) = (0.05M)V titrant (0.1M*2)(0.1L) = (0.05M)V titrant (0.02) = (0.05)V 0.4 = V Multiply the concentration of Ba(OH) 2 by 2 since it will release 2 OH - ions Volume of HCl = 400mL
8. What is the pH of the solution when 50.0mL of 0.010M CH 3 COOH is titrated with 25.0 mL of 0.010M NaOH? K a for acetic acid is 1.8 x 10 -5 .
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9. What is the K sp expression for Pb(OH) 2 ?
10. Calculate the molar solubility of Pb(OH) 2 given that K sp =1.2 x 10 -15

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