Wednesday, May 11

We started class today by going over yesterday's blog. We then spent the first half of class doing page 15 in our packets which was about buffers. A buffer is a solution that resists changes in pH value. That means that when you add an acid or a base to a solution, the pH doesn't change. Then we talked abot what is in a buffer. It is either a weak acid with the conjugate base or a weak base with the conjugate acid. If the buffer was a weak acid with the conjugate base, i would have HA and A-. We then talked about why buffers work. They work becuase the acid won't react with the base. When you add the base, the acid neutralizes so it reacts away what you add. Because of this, the pH doesn't change. The most common buffer is our blood becuase for us to stay alive, the pH has to stay between 7 and 7.8. Most of this is explained in the paragraph at the top of page 15.

We then did problem 28 which was asking if sets of compounds would create a solution that is a buffer. In the first box, the first answer was yes because HNO2 is a weak acid. The other compound listed NaNO2 is a salt containing the conjugate base, NO2-, so it is buffer solution. The next answer is no because HNO3 is a strong acid and a buffer solution has to have a weak acid or weak base, not a strong one. The next solution was a buffer because it has a weak acid with a salt cotaining the conjugate base. The last one in that box is not a buffer because Ca(OH)2 is a strong base. Using those explanations, we did the second box. The answers are yes, yes, no, no.

Next we did problem 29 a. We had to start by wrinting the equation which was H2O+H2CO3<-->HCO3-+H3O+. We were told the initial concentrations of H2CO3 and HCO3- so we put those vlaues in our ICE chart. After adding and subtracting the x's in the middle row of the ICE chart, the bottom row showed the concentrations which were .1-x for H2CO3, .25+x for HCO3- and x for H3O+. We were given the Ka vlaue, 4.4e-7, so we had to set that equal to [HCO3-][H3O+]/[H2CO3]. By plugging in the vlaues we got in the bottom row of the ICE chart, the equation now looks like, 4.4e-7=x(.25+x)/.1-x. By using the 5 percent rule, we can make the problem simpler by eliminating the x's. The equation is now 4.4e-7=.25x/.1. After solving, x=1.76e-7. The problem is asking for pH so we have to recognize that x=[H3O+]. We can then use the equation pH=-log[H3O+] to find that the pH=6.75. After doing this problem we did 29 b on our own. The x value was 2.88e-5. Since this was a base problem, x=[OH-] so we had one more step than we had in the previous problem. Using [OH-] we found the pOH, 4.54, and had to subtract that from 14 to get pH=9.46.

Then we finished doing trials of our lab which was explained in the previous blog. Our homework was to finish the webassign due Friday.