Assume you have made a buffer using Acetic Acid and NaOH, and the pH is 5.74. Using the literature pKa value for acetic acid (i.e. 4.74), calculate the ratio of [OAc-]/[HOAc] in the solution.
Buffers work by resisting pH changes when strong acids or bases are added to them. They do this by absorbing the additional H+ or OH- changes that occur when adding these strong acids/bases, therefore minimizing the change in pH. Henderson Hasselbalch: pH = pka + log(A-/HA) ---> (This is the log of the base concentration divided by acid concentration) Acetic acid is a weak acid. This means that it doesn't completely deprotonate when placed in water, like HCl (a very strong acid) does. At the pKa, half of the molecules are protonated (acid form) and half of the molecules are deprotonated (base form). Just as an example, lets say we make a buffer where the pH is 4 and the pka of the weak acid in our buffer is also 4. From the H.H. equation we get: 4 = 4 + log(x) log(x) = 0 x = 1 . Remember that "x" represents (A-/HA). So the A-/HA = 1, which means that there must be equal amounts of A- and HA. So, at the pKa, there's an equal amount of the base form as there is of the acid form. To solve this question, we can use the H.H. equation to figure out about how much of the buffer is in the acid vs base form at this point. 5.74 = 4.74 + log(x) Now we must solve for x. log(x) = 1 x = 10. Remember that the X represents (A-/HA), so the ratio of base to acid is about 10/1 or 10:1 So the ratio of [OAc-]/[HOAc] is about 10:1
In the presence of a competitive inhibitor, how do Km(apparent) and Vmax change for a given enzyme?
Enzymes catalyze reactions. They bind the substrates, and then facilitate a reaction by decreasing the activation energy. A competitive inhibitor binds to free enzyme (not enzyme-substrate complex, as seen in uncompetitive inhibition). The Vmax is the maximum velocity a specific concentration of enzymes can operate. The Km is the substrate concentration at which the enzyme is operating at half of its maximum velocity (1/2 Vmax). So, if we add competitive inhibitors to our solution of enzymes, they begin binding to free enzyme. When they are bound, the enzyme cannot catalyze reactions. So, now there are fewer enzymes available to catalyze reactions. But, if we keep adding more and more substrate, these inhibitors will actually become out competed, and Vmax can still be reached. Again, in order for this to happen, we have to add LOTS of substrate. So, the Vmax will not change overall. The Km, however, tells us the substrate concentration at which the enzyme is operating at half of its maximum velocity (1/2 Vmax). We just established that we have to add LOTS of substrate to get our enzymes to operate at the maximum velocity, so it makes sense that the amount of substrate we must add to get the enzymes to half Vmax is also increased. From this we get: Vmax stays the same, Km(apparent) increases. The reason for the "(apparent)" is that technically the Km for any given enzyme doesn't change. However, adding inhibitors makes it LOOK like it changes, since we have to add so much more substrate to reach 1/2 Vmax. That is why we call it "apparent" Km.
Find the x and y intercepts for the following equation: 8y - 4x = 16
This question may look scary, especially since our equation is not in the familiar, y=mx+b form. However, all we need to know is, what is an "intercept"? An intercept refers to where exactly the graph crosses the axes. We also know that at the x axis, y always equals 0. At the y axis, x always equals zero. So, if we set y=0 in the above equation, we can find out what the x value is when y=0, or when the graph intercepts the x axis. 8(0) - 4(x) = 16 . Anything multiplied by 0 is always 0. -4(x) = 16 . Divide both sides by -4. x = -4 . So, at x=-4, the graph intercepts the x axis. We can do the same thing to find the y intercept. The intercept occurs when x=0. 8y - 4(0) = 16 Anything multiplied by 0 is always 0. 8y = 16 . Divide both sides by 8. y=2 . So, at y=2, the graph intercepts the y axis.