Respuesta :
Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
When The Molar solubility of AgBr is = 51.33 × 10⁻¹³
Calculation of Solubility of AgBr
Given as per question:
The Amount of NaBr is = 0.150 M
Then Ksp (AgBr) is = 7.7 × 10⁻¹³
Now we Find:
The Molar solubility of AgBr
The we Computation is:
The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr
After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150
Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³
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