Calculate the [H_3O^+] for the following conditions pH = 4.71 pOH = 1.57 Draw th

Question

Calculate the [H_3O^+] for the following conditions pH = 4.71 pOH = 1.57 Draw the following peptide, including all formal charges and structure, at pH 2 pH 8 pH 11 gly-leu-his-cys-asp-tyr Calculate the pH of the following conditions. HCOOH has a pKa = 3.75. pH of a 3.9 x 10^-1 M HCOOH solution 6.2 x 10^-2 M HCOOH and 2.6 x 10^-2 M NaCOOH. HCOOH has a pKa = 3 75. The enzymology lab has a lot of variability in the data. What changes would you suggest that could make the data more consistent? Consider the protocol, the tools used, quantities, volumes, basically everything you can think of the lab. What is the reason for the bell shaped curve of velocity as pH was changed from 1.5 to 13? You titrate an unknown; dissolved it, acidified the solution with an excess of HCI, then titrated with NaOH until a pH of 13.5 was reached. Use the following information to determine; The molecular weight of the unknown Calculate the percent error from your determined amino acid

Explanation / Answer

1)
a)
Ph = 4.71
Ph = -log[H3O+]
4.71 = -log[H3O+]
[H3O+] = 10^(-4.71)
[H3O+] = 1.95*10^-5

b)

Poh = 1.57
Ph = 14- Poh
Ph = 14 - 1.57
Ph = 12.43

Ph = -log[H3O+]
12.43 = -log[H3O+]
[H3O+] = 10^(-12.43)
[H3O+] = 3.72*10^-13

3)
a)
given C = 3.9*10^-1

   Pka = 3.75
-logKa = 3.75
     Ka = 10^(-3.75)
     Ka = 1.77*10^-4
we know the formula
[H+] = [Ka*C]^1/2
[H+] = [1.77*10^-4 *3.9*10^-1]^1/2
[H+] = 8.3*10^-3
Ph = -log[H+]
Ph = -log[8.3*10^-3]
Ph = 2.08

b)

Ph = Pka + log {[salt]/[acid]}

given
   Pka = 3.75
   [salt] = 2.6*10^-2
   [acid] = 6.2*10^-2

Ph = 3.75 + log[2.6*10^-2/6.2*10^-2]
Ph = 3.75 -0.377
Ph = 3.37

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