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
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
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
Option c water is non polar and can dissolve non polar compounds such as oil water is unable to dissolve or mix with substances like oil therefore making the statement false
