A 100 ml test portion from a mining site requires 9.62 mL of 0.000169M Ag+ titrant to reach the endpoint using the reaction described above. What is the CN^-1 concentration in this water sample?

Thank you.

1 Answer
Apr 18, 2016

Answer:

The concentration of #"CN"^"⁻"# is 423 mg/L.

Explanation:

This appears to be a titration of cyanide ion with silver nitrate solution.

You use a 5-(p-dimethylaminobenzylidine)rhodanine indicator and titrate to the first colour change from yellowish-brown to a pink ("salmon") colour.

The equation for the reaction is

#"Ag"^+"(aq) + CN"^"⁻" "(aq)" → "AgCN(s)"#

The conversions involved are

#"Volume of Ag"^+ → "moles of Ag"^+ → "moles of CN"^"⁻" → "mass of CN"^"⁻" → "concentration of CN"^"⁻"#

1. Moles of #"Ag"^+#

#"Moles of Ag"^+ = 9.62 color(red)(cancel(color(black)("mL Ag"^+))) × ("0.169 mmol Ag"^+)/(1 color(red)(cancel(color(black)("mL Ag"^+)))) = "1.626 mmol Ag"^+#

2. Moles of # "CN"^"⁻"#

#"Moles of CN"^"⁻" = 1.626 color(red)(cancel(color(black)("mmol Ag"^+))) × ("1 mmol CN"^"⁻")/(1 color(red)(cancel(color(black)("mmol Ag"^+)))) = "1.626 mmol CN"^"⁻"#

3. Mass of #"CN"^"⁻"#

#"Mass of CN"^"⁻" = 1.626 color(red)(cancel(color(black)("mmol CN"^"⁻"))) ×("26.02 mg CN"^"⁻")/(1color(red)(cancel(color(black)("mmol CN"^"⁻")))) ="42.30 mg CN"^"⁻"#

4. Concentration of #"CN"^"⁻"# in test portion

#["CN"^"⁻"] = "42.30 mg"/"0.100 L" = "423 mg/L"#