Laboratory Testing & General Mineral Processing Engineering

Laboratory Testing & General Mineral Processing Engineering 2017-04-04T06:57:51+00:00
  • To participate in the 911Metallurgist Forums, be sure to JOINLOGIN
  • Use Add New Topic to ask a New Question/Discussion about Mineral Processing or Laboratory Work.
  • OR Select a Topic that Interests you.
  • Use Add Reply = to Reply/Participate in a Topic/Discussion (most frequent).
    Using Add Reply allows you to Attach Images or PDF files and provide a more complete input.
  • Use Add Comment = to comment on someone else’s Reply in an already active Topic/Discussion.

Preparation of Silver Nitrate Solution (7 replies and 2 comments)

Maya Rothman
2 years ago
Maya Rothman 2 years ago

Please, how can one prepare say a 0.1 Molar Silver Nitrate Solution to be used in the titration of Cyanide. Please back it with reactions and or calculations if there are any. Thanks.

Alan Carter
2 years ago
Alan Carter 2 years ago


The Preparation recipe of Silver Nitrate Solution goes as follow:

* g AgNO3/1 H2O desionizada...
1 ml consumido de AgNO3 = 10 ppm NaCN
16.9874 Silver Nitrate AR in 1000ml distilled water.
Test water for chlorides first.

Bob Mathias
2 years ago
Bob Mathias 2 years ago

Must be stored in a dark bottle otherwise it goes off quickly. Decomposes with light. Make up fresh solutions regularly. First determine the Molar mass for Silver Nitrate which is 169.874 g/mol, and then calculate the mass of Silver Nitrate which must added into the Volumetric flask using the following formula
C =m/MV
: m = CMV
: m = (0.1M)(169.87 g/mol)(1L) (That's if you are using a 1 Litre volumetric flask) : = 16.987 g/L

Bill Fraser
2 years ago
Bill Fraser 2 years ago

In addition to the above and FYI: AgNO3 + 2NaCN = NaAg(CN)2 + NaNO3

Assume 50-ml sample aliquot for titration & 1-mL 0.1M AgNO3 requirement:
From eqn: 1000 mL M AgNO3 = 2*49-g NaCN
Or 1-mL 0.1M AgNO3 = 0.0098-g NaCN
In 50-mL aliquot = 196-mg/L NaCN or 104-mg/L free CN

Paul Morrow
2 years ago
Paul Morrow 2 years ago

if you need to measure the amount of cyanide in a solution is more easy if use a solution with 0.4333 grs of AgNO3 in 1L distilled water, this is equivalent to 10 ppm of CN

Bill Fraser
2 years ago

Paul, 10-ppm NaCN (NOT FREE CN) in a 25-mL sample aliquot...5.3-ppm Free CN!

Paul Morrow
2 years ago


Tony Verdeschi
2 years ago
Tony Verdeschi 2 years ago

The actual concentration of Silver Nitrate we use for NaCN titration is 0.0102M ie. you've to take 1.733g AgNO3/H20 Litre but if you want to prepare 0.10M of AgNO3 then you need to take 16.9901g AgNO3/H2O Litre.

2 years ago
David 2 years ago
2 years ago
David 2 years ago

The use of 0.1 M AgNO3 solution is not needed. Page 33 from Dorr's Cyanidation & Concentration of Au & Ag Ores

Standard silver nitrate solution is 4.33 gm of AgNO3 in 1 liter of distilled water. The reaction is AgNO3 + 2NaCN = NaAg(CN)2 + NaNO3. One mole of AgNO3 (169.9 gm) reacts with 2 moles NaCN (98 gm) or 4.33 gm AgNO3 reacts with 2.5 gm NaCN.

Thus a 25 ml sample of solution for titration will require 1 ml of AgNO3 solution per 10 ppm of NaCN contained in the 25 ml sample.

Please join and login to participate and leave a comment.

BUY Laboratory & Small Plant Process Equipment

We have all the laboratory and plant equipment you need to test or build/operate your plant.

ENTER our Mining Equipment' Store

We Sell EQUIPMENT for all types of Mineral Treatment PROCESSES and Laboratory Testing needs

Have a Mineral Processing QUESTION?

Come in, ask your question

911Metallurgist Community Forums

Talk to other metallurgists and be helped.


We can IMPROVE ALL PLANTS / Mineral Processing Engineering & LABORATORY Ore Testing

911Metallurgy Engineering

Contact us for process engineering, metallurgical investigations, plant optimization, plant troubleshooting, needs. WE “FIX” METALLURGY.