Question

When 25.0 mL of a solution of 1.4 x 10-3 M silver nitrate is mixed with 60.0 mL of a solution of 7.5 x 10-4 M sodium chloride a) What is the molar concentration of silver ions in the final mixture? b) What is the molar concentration of chloride ions in the final mixture? Ksp (AgCl) = 1.8 x 10-10

Answer 1

Step  one write  the  equation  for  dissociation  of  AgNO3  and  NaCl

that  is   AgNO3-------> Ag+   + NO3-

            NaCl-------->   Na+   +  Cl-
then  find  the  number  of  moles  of  each  compound
that  is  for   AgNO3  = ( 1.4 x10^-3 ) x  25/1000=  3.5  x10^-5  moles
                     Nacl=  (7.5  x10^-4)x 60/1000=  4.5  x10^-5  moles

from   mole  ratio  the  moles  of    Ag+=  3.5  x10^-5 moles  and  that  of  Cl-= 4.5  x10^-4 moles

  then find  the  total    volume  of  the  mixture
that  is  25ml  +  60 Ml  =85ml  =  0.085 liters

The  Ksp  of  Agcl =  (Ag+) (cl-),    let  the  concentration  of  Ag+ be  represented   by  x  and   also  the  concentration  be  represented by  x

ksp (1.8 x10^-10)  is  therefore=  x^2

find  the  square  root   x=1.342   x10^-5

Ag+ in  final  mixture  is =  moles  of  Ag+/total  volume - x
that is  {(3.5  x10^-5)/0.085}  -  1.342  x10^-5=3.98x10^-4


Cl-  in  the  final  mixture   is =(4.5  x10^-5 /0.085)  -  1.342  x10^-5= 5.16 x10^-4

Answer 2

First, we need to get  moles of AgNO3:
moles of AgNO3 = molarity * volume
                             = 1.4x10^-3 M* 0.025 L
                             = 3.5x10^-5 moles

then, moles of NaCl = molarity * volume
                                  = 7.5x10^-4  M * 0.06 L
                                  = 4.5 x 10^-5 moles
by using this equation:
AgCl → Ag+   +  Cl-
Ksp = [Ag+][Cl-]
when we have Ksp (AgCl) = 1.8x10^-10
and by assuming [Ag+] & [Cl-] = X so,by substitution:
1.8x10^-10 = X*X
X^2 = 1.8x10^-10
∴X = 1.3 x 10^-5
[Ag+] = [Cl-] = 1.3x10^-5 M 

and when the total final volume = 0.025 L + 0.06 L = 0.085 L
∴ [Ag+] in the final mixture = ((3.5x10^-5) - (1.3x10^-5))/0.085L
                                             = 2.5 x 10^-4  M
∴[Cl-] in the final mixture = ((4.5x10^-5) - (1.3x10^-5))/ 0.085 L 
                                          = 3.8 x 10^-4 M