A sample of NaCl weighing 0.2286g is dissolved in H2O and exactly 50.0mL of silver nitrate solution is added to precipitate AgCl. The excess Ag+ is titrated with 12.55mL of 0.0986N solution of KSCN. What is the normality of the AgNO3 solution?

1 Answer
Ernest Z.
Feb 14, 2018

The normal concentration of the #"AgNO"_3# is 0.103 equiv/L.

Explanation:

A. The reaction with #"NaCl"#

The equation for the reaction is

#"AgNO"_3 + "NaCl → AgCl + NaNO"_3#

Step 1. Calculate the equivalents of #"NaCl"#

#"Equiv. of NaCl" = 0.2286 color(red)(cancel(color(black)("g NaCl"))) × "1 equiv. NaCl"/(58.44 color(red)(cancel(color(black)("g NaCl")))) ="0.003 912 equiv. NaCl"#

Step 2. Calculate the equivalents of #"AgNO"_3#

#"Equiv. of AgNO"_3 = "0.003 912" color(red)(cancel(color(black)("equiv. NaCl"))) × "1 equiv. AgNO"_3/(1 color(red)(cancel(color(black)("equiv. NaCl")))) ="0.003 912 equiv. AgNO"_3#

B. The reaction with #"KSCN"#

The equation for the reaction is

#"AgNO"_3 + "KSCN → AgSCN + KNO"_3#

Step 3. Calculate the equivalents of #"KSCN"#

#"Equiv. of KSCN" = "0.012 55" color(red)(cancel(color(black)("L KSCN"))) × "0.0986 equiv. KSCN"/(1 color(red)(cancel(color(black)("L KSCN")))) ="0.001 237 equiv. KSCN"#

Step 4. Calculate the equivalents of #"AgNO"_3#

#"Equiv. of AgNO"_3 = "0.001 237" color(red)(cancel(color(black)("equiv. KSCN"))) × "1 equiv. AgNO"_3/(1 color(red)(cancel(color(black)("equiv. KSCN")))) ="0.001 237 equiv. AgNO"_3#

Step 5. Calculate the total equivalents of #"AgNO"_3#

#"Total equiv of AgNO"_3 = "(0.003 912 + 0.001 237) equiv AgNO"_3 = "0.005 149 equiv. AgNO"_3#

Step 6. Calculate the normal concentration of #"AgNO"_3#

#N = "equivalents"/"litres" = "0.005 149 equiv."/"0.0500 L" = "0.103 equiv/L"#

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