Question

calculate the percent ionization of a 0.125 m solution of nitrous acid (a weak acid), with a ph of 2.09. (answer should have two significant figures, expressed in %)

Answer 1

The percent ionization of a 0.125 M solution of nitrous acid with a pH of 2.09 is approximately 970%.

Step-by-step explanation:

The percent ionization for an acid can be calculated using the formula:

Percent Ionization = antilog(pH) / initial concentration * 100%

Given that the pH is 2.09 and the initial concentration is 0.125 M, we can calculate the percent ionization as follows:

Percent Ionization = antilog(2.09) / 0.125 * 100% = 1.21 / 0.125 * 100% = 969.6%

Therefore, the percent ionization of the 0.125 M solution of nitrous acid with a pH of 2.09 is approximately 970% (rounded to two significant figures).

Answer 2

To find the percent ionization of nitrous acid with a pH of 2.09, convert the pH to the hydrogen ion concentration and then divide by the initial concentration of the acid. The calculation yields approximately 6.5% ionization.

Step-by-step explanation:

To calculate the percent ionization of a 0.125 M solution of nitrous acid (HNO2) with a pH of 2.09, we must first convert the pH into the concentration of hydrogen ions ([H+]) using the formula [H+] = 10-pH. In this case, [H+] = 10-2.09 = 8.13 × 10-3 M. Since nitrous acid is a weak acid, not all of it will ionize. The percent ionization is then found by dividing the concentration of ionized acid by the initial concentration of acid and multiplying by 100 to get a percentage: percent ionization = ([H+]/[HNO2 initial]) × 100. Inserting the calculated [H+] and the given initial concentration of nitrous acid, 0.125 M, we find the percent ionization: percent ionization = (8.13 × 10-3/0.125) × 100 ≈ 6.5%.

Remember that the provided pH value is logarithmic, and the result is limited to two significant digits.