If the solubility product of AgBrO3 and Ag2SO4 are 5.5 x 10^-5 and 2 x 10^-5 respectively. The relationship between the solubilities of these can be correctly represented as
Correct answer: A. sAgBrO3 > sAg2SO4
- A. sAgBrO3 > sAg2SO4
- B. sAgBrO3 < sAg2SO4
- C. sAgBrO3 = sAg2SO4
- D. sAgBrO3 << sAg2SO4
Explanation
The greater the solubility product higher the solubility. To determine the relationship between the solubilities of AgBrO3 and Ag2SO4 based on their solubility product constants (Ksp), we can compare the magnitudes of the Ksp values. Given that the Ksp of AgBrO3 is 5.5 x 10-5 and the Ksp of Ag2SO4 is 2 x 10-5, we can compare these values to determine their relative solubilities. If we divide, we get: (5.5 x 10-5) / (2 x 10-5) = 2.75 This tells us that the Ksp of AgBrO3 is 2.75 times larger than the Ksp of Ag2SO4. Since the solubility product constant (Ksp) is related to the solubility of a compound, we can infer that AgBrO3 is more soluble than Ag2SO4.
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