AgNO3 1.2243 g 500.0 mL Ag^+ 25.00 mL 37.38 mL Cl^-

AgNO3 1.2243 g 500.0 mL Ag^+ 25.00 mL 37.38 mL Cl^-

 

 

a) Standard Ag^+ solution was prepared by dissolving

1.2243 g of dry AgNO3 (FW 169.87) in water in a 500.0 mL volumetric flask.

A dilution was made by delivering 25.00 mL of solution

with a pipet to a second 500.0 mL volumetric flask and diluting to the mark.

Find the concentration of Ag^+ in the dilute solution.

 

b) A 25.00 mL aliquot of unknown containing Cl^-

was titrated with the dilute Ag^+ solution,

and the equivalent point was reached

when 37.38 mL of Ag^+ solution had been delivered.

Find the concentration of Cl^- in the unknown.

 

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a)

1.2243 g / (169.87 g/mol) = 0.0072073 mol AgNO3

( 참고 https://ywpop.tistory.com/7738 )

 

= 0.0072073 mol Ag^+

 

 

 

0.0072073 mol / (500.0/1000 L) × (25.00/1000 L) / (500.0/1000 L)

= 0.0072073 / (500.0/1000) × (25.00/1000) / (500.0/1000)

= 0.00072073 mol/L

= 7.207×10^(-4) M Ag^+

 

 

 

b)

MV = M’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(7.207×10^(-4) M) (37.38 mL) = (? M) (25.00 mL)

 

? = (7.207×10^(-4)) (37.38) / (25.00)

= 0.001078 M

= 1.078×10^(-3) M Cl^-