Respuesta :
There is 0.00025 moles of NaBrO₃ in the 0.5 L solution (0.5L×0.00050M=0.00025mol) which means that when 0.0010 moles of AgNO₃ is added only 0.00025 moles of AgBrO₃ can be made. In other words, NaBrO₃ is the limiting reagent and since NaBrO₃ reacts with AgNO₃ in a 1 to 1 molar ratio, we know that only 0.00025 AgBrO₃ will be made before the amount of NaBrO₃ presant is used up. I am assuming that the volume of the solution does not change while AgNO₃ is being added since the question does not state it which means that the concentration of AgBrO₃ is 0.0005M (0.00025mol÷0.5L=0.0005M)
To find out if AgBrO₃ precipitates out of solution you need to find the molar solubility of AgBrO₃ which is found by setting up the equation 5.4×10⁻⁵=[x][[x] and solving for x. (this eqation basically the equation for Ksp but replacing the solute concentrations with the variable x). When doing this you find that x=0.0073 mol/L which means that 0.0073 moles of AgBrO₃ can dissolve in 1 L of solution. That means that 0.00367moles can dissolve in 0.5 L of solution.
The first part of the question we found that we could produce 0.00025 moles of AgBrO₃ in the 0.5 L solution and in part two we found that 0.00367moles of AgBrO₃ in 0.5 L solution which means that no precipitate will form.
I hope this helps. Let me know if anything is unclear or doesn't make sense. I have not done a problem like this in a while so there could be some mistakes so I encourage you to check the work and see if it makes sense to you.
To find out if AgBrO₃ precipitates out of solution you need to find the molar solubility of AgBrO₃ which is found by setting up the equation 5.4×10⁻⁵=[x][[x] and solving for x. (this eqation basically the equation for Ksp but replacing the solute concentrations with the variable x). When doing this you find that x=0.0073 mol/L which means that 0.0073 moles of AgBrO₃ can dissolve in 1 L of solution. That means that 0.00367moles can dissolve in 0.5 L of solution.
The first part of the question we found that we could produce 0.00025 moles of AgBrO₃ in the 0.5 L solution and in part two we found that 0.00367moles of AgBrO₃ in 0.5 L solution which means that no precipitate will form.
I hope this helps. Let me know if anything is unclear or doesn't make sense. I have not done a problem like this in a while so there could be some mistakes so I encourage you to check the work and see if it makes sense to you.
