Calculate the pH of a 0.1 M HCl solution. Calculate the pH of a 0.1 M NaOH solut

Question

Calculate the pH of a 0.1 M HCl solution. Calculate the pH of a 0.1 M NaOH solution What is the concentration of [H^+] in molars, millimolars, and micromolars for a solution of pH 5? If you mix 10 mL of a 0.1 M HCl solution with 8 mL of a 0.2 M NaOH solution, what will be the resulting pH? If a weak acid, HA. is 3% dissociated in a 0.25 M solution. calculate the K_a and the pH of the solution What is the pH of a 0.05 M solution of TRIS acid (pK_a = 8.3)? What is the pH of a 0.045 M solution of TRIS base? If you mix 50 mL of 0.1 M TRIS acid with 60 mL of 0.2 M TRIS base, what will be the resulting pH? If you add 1 mL of 1 M NaOH to the solution in 8 above, what will be the pH? How many total milliliters of 1 M NaOH can you add to the solution in Problem 8 and still have a good buffer (that is, within 1 pH unit of the pKa)? If you are making 100 mL of a 0.1 M HEPES buffer starting from HEPES in the basic form, is it prudent to get 50 mL of 1 M HCl from the community reagent bottle to use for your titration? An enzyme-catalyzed reaction is earned out in a 150 mL solution containing 0.1 M TRIS buffer. The pH of the reaction mixture at the start was 8.0. As a result of the reaction, 0.002 mol of H+ were produced. What is the ratio of TRIS base to TRIS acid at the start of the experiment? What is the ratio at the end of the experiment? What is the final pH? The pKa of HFPES is 7.55 at 20 degree C, and its MW is 238, 21. Calculate the amounts of HFPES in grams and of 1.0 M NaOH in milliliters that would be needed to make 300 mL of 0.2 M HEPES buffer at pH 7.2.

Explanation / Answer

1) the pH can be calculated as

pH = -log[H+]

pH = -log[0.1] = 1

2) the pOH of NaOH will be

pOH = -log[OH-] = -log[0.1] = 1

So pH = 14 - pOH = 14 - 1 = 13

3) the moles of HCl = Molarity X volume = 0.1 X 10 = 1 millimoles

moles of NaOH = Molarity X volume = 0.2 x 8 = 1.6 millimoles

so 1 millimole of HCL will react with 1 millimoles of NaOH

Millimoles of NaOH left = 1.6 - 1 = 0.6

Total volume = 10 + 8 = 18mL

[NaOH] = 0.6 / 1.8 = 0.33M

pOH = -log0.33 = 0.48

pH = 14 - 0.48 = 13.52

5) For weak acid we know that

Ka = [H+]^2 / [HA]

[H+] = 0.03 X 0.25 = 0.0075

[HA] = 0.25 - 0.0075 = 0.25 (approx)

Ka = 0.0075 X 0.0075 / 0.25 = 0.000225

pH = -log[H+] = -log[0.000225] = 3.65

6) [H+] of tris will be

[H+] = (Ka X concentration)^1/2

pKa = 8.3

Ka = 5.01 X 10^-9

[H+] = (5.01 X 10^-9 x 0.05)1/2

[H+] =( 2.5 X 10^-10)1/2

[H+] = 1.58 X 10^-5

pH = -log[H+] = 4.8

8) pH = pKa + log[base] / [acid]

pH = 8.3 + log [0.2 / 0.1] = 8.6

9) If we will add 1mL of 1M NaOH so moles of naOH added = 1 millimoles

Moles of NaOH will react with tris acid to give salt

So new pH = 8.3 + log [0.2 +0.001 / 0.1-0.001] = 8.6

10) The pH change = 1

Which means the [salt] / [acid] = 10

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