Asked by chem question
                For which of the following mixtures will Ag2SO4(s) (Ksp=1.20E^-5)precipitate?
More than one answer can be selected.
a.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)
b.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)
c.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)
d.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)
            
        More than one answer can be selected.
a.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)
b.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)
c.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)
d.) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)
Answers
                    Answered by
            DrBob222
            
    Calculate (Ag^+) and (SO4^2-), substitute into Ksp expression. That is Qsp. 
If Qsp > Ksp a ppt will occur. If Qsp < Ksp, no ppt will occur.
For example, a)
(SO4^2-) = 150.0 x 0.10/155.0)= 0.0968M
(Ag^+) = 5.0 x 0.2/155.0 = 0.00645M
(0.00645)^2(0.0968) = 4.03E-6 which is smaller than Ksp = 1.20E-5; therefore, no ppt of Ag2SO4 will occur.
E-6
    
If Qsp > Ksp a ppt will occur. If Qsp < Ksp, no ppt will occur.
For example, a)
(SO4^2-) = 150.0 x 0.10/155.0)= 0.0968M
(Ag^+) = 5.0 x 0.2/155.0 = 0.00645M
(0.00645)^2(0.0968) = 4.03E-6 which is smaller than Ksp = 1.20E-5; therefore, no ppt of Ag2SO4 will occur.
E-6
                    Answered by
            chem question
            
    Thanks, I got it!
    
                    Answered by
            Thien
            
    Why do you have to square Ag after you got the Molarity?
-(0.00645)^2(0.0968)=4.03E-6
    
-(0.00645)^2(0.0968)=4.03E-6
                    Answered by
            Juliet
            
    Yes square molarity
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.