Nitric acid is used to etch copper during printmaking. To test the strength of a

Question

Nitric acid is used to etch copper during printmaking. To test the strength of a nitric acid solution, 5.00 mL of a commercial preparation of this acid is dissolved in water and diluted to 100.00 mL in a volumetric flask. A 25.00 mL portion of this diluted acid is titrated with 0.5365 M NaOH and gives and endpoint at 36.87 mL.

a. What is the concentration of the nitric acid in the original commercial preparation?

b. What is the expected pH of the diluted nitric acid sample solution at the beginning of the titration?

c. What is the expected pH during this titration after 10.00 mL, 25.00 mL, and 40.00 mL of titrant have been added?

Explanation / Answer

a. Concentration of HNO3 before titration

Concentration = 0.5365 x 36.87 / 25 = 0.79 M ( Using V1S1 = V2S2)

Agian conectration efore dilution will be ( commercial conecntration)

Cocentration = 105 x 0.79 / 5 = 16.62 M

the concentration of the nitric acid in the original commercial preparation is 16.62 M

b. pH = -log [H+] = - log 16 = 1.2 ( Since concentration will be 16.62 M for [H]

c.

pH after adding 10 mL of NaOH will be

no of mmol of HCl = 25 x 0.79 = 19.75 mmol

and no of mmol of NaOH = 0.5365 x 10 = 5.365 mmol

So, ammount excess HCl added = 19.75 - 5.365 =14.39 mmol

Hence concentration will be 14.39 / (25 + 10 ) = 0.411

pH = -log [H+] = -log 0.411 = 0.39

pH after adding 25 mL of NaOH will be

no of mmol of HCl = 25 x 0.79 = 19.75 mmol

and no of mmol of NaOH = 0.5365 x 25 = 13.41 mmol

So, ammount excess HCl added = 19.75 - 13.41 =6.34 mmol

Hence concentration will be 6.34 / (25 + 25 ) = 0.127

pH = -log [H+] = -log 0.127 = 0.9

pH after adding 40 mL of NaOH will be

no of mmol of HCl = 25 x 0.79 = 19.75 mmol

and no of mmol of NaOH = 0.5365 x 40 = 21.46 mmol

So, ammount excess NaOH added = 21.46 - 19.75 =1.71 mmol

Hence concentration will be 1.71 / (25 + 40 ) = 0.026

pOH = -log [OH-] = -log 0.026 =1.58

So, pH = 14- 1.58 =12.42

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