You wish to create a bufer inrus acid as your weak acid CHNO, K,-724. Part 1: If

Question

You wish to create a bufer inrus acid as your weak acid CHNO, K,-724. Part 1: If you have 41.4 mL of an 0.121 M solution of HN02 and 0.147 M solution of Nor what is the pH? 3.22 Part 2: Supposing you wish to create a solution with a pH of 3.27. If you begin with 41.4 mL of a 0.121M solution of HNO2, how many grams of NaNo, would you have to add to the solution. Assume that the volume does not change in the process. 0.466 Part 3: Instead, you begin with 41.4 mL of a 0.121 M solution of HN02 and decide to create your buffer solution by adding NaOH to the system. How many grams of NaOH would you have to add in order to have your desired pH of 3.27? Assume that the volume does not change in the process. 0.27

Explanation / Answer

Answer

0.1151g

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([A-] /[HA])

3.27 = 3.14 + log([NO2-] /[HNO2])

log([NO2-] /[HNO2]) = 0.13

[NO2-] /[HNO2] = 1.3490

[NO2-] = 1.3490[HNO2]

Buffer concentration = 0.121M

[ HNO2] + [NO2-] = 0.121M

  [HNO2] + 1.3490[HNO2] = 0.121M

   2.3490[HNO2]= 0.121M

   [HNO2] = 0.0515M

[NO2-] = 0.121M - 0.0515M = 0.0695M

Initial moles of HNO2 = (0.121mol/1000ml) × 41.4ml = 0.0050094

No of moles of HNO2 required = (0.0515mol/1000ml)×41.4ml = 0.0021321

No of moles of HNO2 to be neutralized = 0.0050094 - 0.0021321 = 0.0028773

NaOH + HNO2 - - - - - - > NaNO2 + H2O

this is 1:1molar reaction

Therefore

to neutralize 0.0028773mole of HNO2, 0.0028773 moles of NaOH needed

Molar mass of NaOH = 40g/mol

Therefore

Mass of NaOH needed = 40g/mol × 0.0028773mol = 0.1151g

  

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