Chem2080 2011 Prelim2 KEY - Page 1 of 9 Page Chemistry 2080...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
of 9 Page 1 of 9 Chemistry 2080 - Spring 2011 Prelim #2 Tuesday April 12th, 7:30 to 9:00 pm Name: ___________ KEY_______________________ CU ID: ____________________ Last First M.I. Lab: _______________________________ Lab TA Name: __________________ Day Time Instructions: Answer all questions in the spaces provided. You have 90 minutes. You must show all work for full credit. There are 9 pages in this exam, and there is also a separate equation sheet. All temperatures are assumed to be 25°C unless otherwise stated. If you have to solve a quadratic equation , please write it out first in the standard form, such as: ax 2 + bx + c = 0, and then you can use a calculator to solve it directly. The following are not allowed at the exam: mp3 players, iPods and similar devices, headphones or earbuds, cell phones, iPads, Laptop computers, machines that go “Ping”. There are 6 questions; the points for each are indicated below. 1. _____________/15 2. _____________/16 3. _____________/22 4. _____________/18 5. _____________/15 6. _____________/14 Total _____________/100
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
of 9 Page 2 of 9 1) (15 points) (a) Calculate the pH at 25°C of a 1.28x10 -7 M aqueous solution of the strong base Ca(OH) 2 . Please give your answer to two decimal places. (You will need to take into account the autoionization of water). ______________________________ Calcium hydroxide is a strong base, therefore it ionizes essentially completely in aqueous solution: Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) As the solution is so dilute, we cannot ignore the autoionization of water. We therefore have the following equations: Material balance: [Ca 2+ ] = 1.28x10 -7 M . . . . . . . (1) Electroneutrality: 2[Ca 2+ ] + [H 3 O + ] = [OH - ] . . . . . . . (2) And the definition of the ion-product of water: K w = [H 3 O + ][OH - ] = 1.00x10 -14 . . . . . . . (3) Substituting from equations (1) and (3) into (2), we get: 2 (1.28x10 -7 ) + [H 3 O + ] = K w /[H 3 O + ] i.e. [H 3 O + ] 2 + 2.56x10 -7 [H 3 O + ] 1.00x10 -14 = 0 which is a quadratic equation in [H 3 O + ], so we can use the familiar equation to solve this quadratic, whence: 3 O + ] = 3.443x10 -8 (taking the positive root (the negative root is meaningless)) and thus pH = 7.46 ( If the autoionization of water is ignored, the value would be pH = 7.41) (b) What is the pH at 25°C of the solution obtained by mixing 50 mL of 2.0 x10
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

Chem2080 2011 Prelim2 KEY - Page 1 of 9 Page Chemistry 2080...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online