Question

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.

Answer 1

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"`