Question

Помогите пожалуйста решить.Вычислить pH и pOH в растворе азотной кислоты с C(HNO3)=0,01моль/л.

Answer 1

Дано :
C(HNO3)=0,01 моль/л

pH -?
pOH -?

Решение:
HNO3 - сильная кислота и полностью диссоциирует
HNO3=>H(+) + NO3(-)
pC=C(H+)=0,01 моль/л = 1*10^-2 моль/л

pH=-lgC(H+)
pH=(-lg1) + (-lg10^-2)= 2-0=2 - сильнокислая среда

pH+pOH=14
pOH=14-pH
pOH=12

Answer 2

Calculation of pH and pOH in a Nitric Acid Solution

To calculate the pH and pOH in a solution of nitric acid (HNO3) with a concentration of C(HNO3) = 0.01 mol/L, we can use the following formulas:

pH = -log[H+] pOH = -log[OH-]

In this case, we need to determine the concentration of H+ ions in the solution, which is equal to the concentration of nitric acid (HNO3) since it is a strong acid that dissociates completely in water.

Given that the concentration of nitric acid is C(HNO3) = 0.01 mol/L, we can conclude that the concentration of H+ ions is also 0.01 mol/L.

Using the formula for pH, we can calculate:

pH = -log[H+] pH = -log(0.01) pH ≈ 2

Therefore, the pH of the nitric acid solution is approximately 2.

To calculate the pOH, we need to consider that in water, the concentration of H+ ions multiplied by the concentration of OH- ions is equal to 1x10^-14 at 25°C. Since nitric acid is a strong acid, it completely dissociates into H+ and NO3- ions, and there are no OH- ions present in the solution.

Therefore, the concentration of OH- ions is negligible, and we can assume it to be zero. Consequently, the pOH of the nitric acid solution is also zero.

In summary: - The pH of the nitric acid solution is approximately 2. - The pOH of the nitric acid solution is zero.

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