Question

calculate the H and oh concentration for the three buffer on the picture

Part B: Effect of pH on Amylase Activity 1. Prepare 3 spot plates with one drop of iodine solution in each well. 2. Pipet 2.0 mL of buffers with pH values of 4. 7. and 10 into separate test tubes. Also Piper mi of starch into each test tube. 3. Pipet 2.0 mL of enzyme solution into the pH 7 solution. Immediately add two drops of this solution into a well in a spot plate. This is the zero time observation. 4. After two minutes, add two drops of solution to another well of the spot plate, and so on until the solution remains a light yellow. 5. After the routine for this solution is established, repeat the procedure, steps 3-5, with the pH 4 and pH 10 buffer solutions. Carry out the testing on the pH 4 and pH 10 buffers until the reaction is complete or for 10 minutes longer than the pH 7 reaction took to go to completion. 6. Record all observations.

Answer 1

1) Solution with pH 4 buffer

$\\ \ \\ \ [H^+] = 10^{-pH } \\ \ \\ \ [H^+] = 10^{- 4} \ M \\ \ \\ \ [OH^-] = \frac{K_w}{[H^+]} \\ \ \\ \ [OH^-] = \frac{10^{-14} \ M^2}{10^{- 4} \ M} \\ \ \\ \ [OH^-] = 10^{- 10} \ M$

2) Solution with pH 7 buffer

$\\ \ \\ \ [H^+] = 10^{-pH } \\ \ \\ \ [H^+] = 10^{- 7} \ M \\ \ \\ \ [OH^-] = \frac{K_w}{[H^+]} \\ \ \\ \ [OH^-] = \frac{10^{-14} \ M^2}{10^{- 7} \ M} \\ \ \\ \ [OH^-] = 10^{- 7} \ M$

3) Solution with pH 10 buffer

$\\ \ \\ \ [H^+] = 10^{-pH } \\ \ \\ \ [H^+] = 10^{- 10} \ M \\ \ \\ \ [OH^-] = \frac{K_w}{[H^+]} \\ \ \\ \ [OH^-] = \frac{10^{-14} \ M^2}{10^{- 10} \ M} \\ \ \\ \ [OH^-] = 10^{- 4} \ M$