Question

Acids, Bases and Electrolytes Questions 1Calculate the pH of a 0.10 M nitric acid solution, How does your calculation compare to your experimental pH? 2. Calculate the pOH, pH, and [H1 of a 0.10M KOH solution. How does the pH compare with your data? 3. Use your date to calculate the [H]. pOH, and [OH of the 0.10 M acetic acid solution. 4. Use your data to calculate the pOH, pH, and [H] of the 0.10M ammonia solution. made by mixing 15.0 mL of 0.10 M HNO, with 69.0 mL of A nitric acid solution was 5. water Calculate the molarity of the diluted nitric acid solution. a. b. Calculate [H], pH, pOH, and [OH ] of the diluted solution.

Answer 1

1.

pH = -log [H⁺]

here [H⁺] = concentration of H⁺ ions.

HNO₃ is a strong acid. It is assumed that all HNO3 molecules ionizes and produce equal amount of H⁺ ions. So, the concentration of HNO₃ is equal to concentration of H⁺ ions.   

So, [H⁺] = [HNO₃] = 0.1 M

pH = -log[H⁺] = -log 0.1 = 1

pH of 0.1M HNO3 solution is 1.

2.

KOH is strong base and it is assumed that all KOH molecules are ionizes in solution and equal amount of OH^- ions are produced.

So, [OH^-] = [KOH] = 0.1 M

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

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pOH + pH = 14

pH = 14 - pOH = 14-1 = 13

pH of 0.1 m KOH is 13.

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pH = -log [H⁺]

13 = -log [H⁺]

antilog -13 = [H⁺]

1 x 10^-13 = [H⁺]

[H⁺] of 0.1 M KOH is 1 x 10^-13

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5.

M1V1 = M2V2

M1 = intial molarity and V1 = intial volume

M2 = final molarity and V2 = final volume

M1 = 0.1, V1 = 15.0 mL

M2 = ? , V2 = total vaolume after dilution = 69.0 + 15 = 84 mL

M2 = M1V1 / V2 = 0.1 x 15 / 84 = 0.0178 M.

Molarity of HNO3 solution after dilution is 0.0178 M

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HNO₃ is a strong acid. It is assumed that all HNO3 molecules ionizes and produce equal amount of H⁺ ions. So, the concentration of HNO₃ is equal to concentration of H⁺ ions.   

So, [H⁺] = [HNO₃] = 0. 0178 M

pH = -log[H⁺] = -log 0.0178 = 1.749

pH of 0.0178 M HNO3 solution is 1.749

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pOH + pH = 14

pH = 14 - pOH = 14-1.749 = 12.251

pH of 0.0178 M of HNO3  is 12.251.

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pOH = -log [OH^-]

12.251 = - log[OH^-]

antilog - 12.251 = [OH^-]

[OH^-] = 5.61 X 10^-13