Question

Compare pH in 0.02 M of HNO3 and 0.02 M HNO2 solutions by calculating. Ka (HNO2) = 6.9×10^-4 .​

Answer 1

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Answer 2

Answer:

The answer to this question is 2.47.

Explanation:

The pH of a 0.02 M solution of HNO3 (nitric acid) is lower than the pH of a 0.02 M solution of HNO2 (nitrous acid).

The pH of a solution can be calculated using the expression:

pH = -log[H+], where [H+] is the concentration of hydrogen ions in the solution.

In the case of 0.02 M HNO3, the hydrogen ions are produced by the dissociation of HNO3 into H+ and NO3-. The concentration of H+ ions is equal to the initial concentration of HNO3, which is 0.02 M. Therefore, the pH of 0.02 M HNO3 can be calculated as:

$pH = -log(0.02) = 1.7$

In the case of 0.02 M HNO2, the hydrogen ions are produced by the dissociation of HNO2 into H+ and NO2-. The dissociation constant (Ka) for HNO2 is given as 6.9 x 10^-4. We can use this value to calculate the concentration of H+ ions in the solution:

$[H+] = Ka / [HNO2] = 6.9 x 10^-4 / 0.02 = 3.45 x 10^-3 M$

The pH of 0.02 M HNO2 can be calculated as:

$pH = -log(3.45 x 10^-3) = 2.47$

Therefore, the pH of 0.02 M HNO3 (1.7) is lower than the pH of 0.02 M HNO2 (2.47), meaning that HNO3 is a stronger acid than HNO2. This is due to the fact that HNO3 is a stronger proton donor, meaning it donates more hydrogen ions to the solution than HNO2.

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