I've just learned that when carrying out the identification test for thiocyanate ions borax can induce a false positive and therefore needs to be removed. I couldn't find the answer on google so I'm asking a question here.

I would think you first carry out the test to see if borax is present; but if it is, how would you remove it?

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    $\begingroup$ There are numerous methods of identifying thiocyanate anions both quantitatively and qualitatively. Which one are you asking about? The formation of iron(III) complex? $\endgroup$
    – andselisk
    May 5, 2019 at 13:51
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    $\begingroup$ @andselisk yes. Normally there should be a complex with iron(III), thiocyanate ions and water but with Borax ions present the iron(III) makes a complex with borax. How would one then remove the borax ions? $\endgroup$
    – bobsburger
    May 5, 2019 at 14:07
  • $\begingroup$ Add more context to your question. Which test are you talking about, since this is not a generic interference. Check these tests, it may not need borate removal. books.google.com/… $\endgroup$
    – AChem
    May 5, 2019 at 15:14

2 Answers 2


Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated.

From [1, p. 358]:

6 . Nachweis als $\ce{Fe(SCN)3}$


Störungen: Um den störenden Einfluss von $\ce{F-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{H3BO3}$, $\ce{C4H4O6^2-}$, $\ce{C2O4^2-}$ usw., die mit $\ce{Fe3+}$-Ionen Komplexe bilden, auszuschalten, wird $\ce{Fe^3+}$ im Überschuss zuge­geben.

From [1, p. 418–419] (emphasis mine):

9. Nachweis von Fe(III) als $\ce{Fe(SCN)3}$

$$\ce{Fe^3+ + 3 SCN- <=> Fe(SCN)3}$$


Störungen: […] Ferner beeinträchtigen $\ce{Hg^2+}$-Ionen durch Bildung von wenig dissoziiertem $\ce{Hg(SCN)2}$, $\ce{F-}$ durch $\ce{[FeF6]^3-}$-Komplexbildung, die Anionen organischer Säuren ebenfalls durch Komplexbildung und auch $\ce{PO4^3-}$, $\ce{AsO4^3-}$, Borat-Ionen sowie ein größerer Mineralsäureüberschuss die Reaktion. Es ist daher ratsam, vor der Prüfung […] die störenden Anionen in neutraler Lösung mit $\ce{Ba^2+}$ abzutrennen.


  1. Jander, G.; Blasius, E.; Strähle, J.; Schweda, E.; Rossi, R. Lehrbuch der analytischen und präparativen anorganischen Chemie, 16. Auflage.; Hirzel: Stuttgart, 2006. ISBN 978-3-7776-1388-8. (in German)
  • 1
    $\begingroup$ Seems like a lovely book. Is it for undergraduate level? $\endgroup$
    – AChem
    May 5, 2019 at 16:48
  • $\begingroup$ @M.Farooq Pretty much. Bare-bone analytical/inorganic chemistry, that's why I like it:) $\endgroup$
    – andselisk
    May 5, 2019 at 16:55
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    $\begingroup$ I am just finishing an educational article on the reliability of German translations which are translated by free software. Your reference just reinforced my conviction that as scientists we should use and learn at least one more language. Passing a reading test in Russian or French or German was compulsory, decades ago, for a PhD in chemistry. $\endgroup$
    – AChem
    May 5, 2019 at 17:12

When you ask how to remove borax, I assume it is Sodium tetraborate decahydrate (Wikipedia). Thus, I agree with @andselisk of using $\ce{Ba^2+}$ solution to precipitate it. Even though it is seemingly an excellent answer, the reference given and the text body are in German, and I didn't understand it much. :-) Thus, I want to give some clues to make it sense.

According to Ref.1, tetraborate-ions ($\ce{B4O7^2-}$) form a white precipitate of barium metaborate $\left(\ce{Ba(BO2)2}\right)$ with $\ce{Ba^2+}$ solution ($\ce{BaCl2}$).

However, if it is a neutral solution, only following reaction occurs: $$\ce{B4O7^2- (aq) + Ba^2+ (aq) + 3H2O (l) -> Ba(BO2)2 (s) + 3 H3BO3 (aq)}$$

Meaning, removal of borax is only partially completed. Complete precipitation of $\ce{B4O7^2-}$ ion would be reached only in the strong alkaline medium: $$\ce{2 H3BO3 (aq) + Ba^2+ (aq) + 2 OH- (aq) -> Ba(BO2)2 (s) + 4 H2O (l)}$$

Keep in mind that this white precipitate is dissolved in inorganic acids: $$\ce{Ba(BO2)2 (s) + 2 H3O+ (aq) -> Ba^2+ (aq) + 2 H3BO3 (aq)}$$

Thus, white precipitate of $\ce{Ba(BO2)2}$ should be filtered off before adding $\ce{Fe^3+}$ solution.

Alternatively, I believe testing of thiocyanate ions can be done in presence of borax by doing two back to back reactions: (i) Check the solution with the blood-red color formation reaction with $\ce{Fe^3+}$ solution as usual; and then (ii) Check again the same solution with the blue color formation reaction $\left(\ce{(NH4)2[Co(SCN)4] (aq)}\right)$ with $\ce{Co^2+}$ solution:

$$\ce{Co^2+ (aq) + 4 SCN- (aq) + 2 NH4+ (aq) <=> (NH4)2[Co(SCN)4] (aq)}$$


  1. O. A. Ievtifieieva, V. V. Bolotov, T. A. Kostina, O. M. Svechnikova, T. I. Yuschenko, N. I. Kaminska, A. E. Kosareva, L. V. Slobodyanyuk, O. P. Yashchuk, In Analytical chemistry (Qualitative analysis) Part I: The manual for students of higher schools; O. A. Ievtifieieva, Ed.; Publishing house the CLL: Kharkiv, Ukraine, 2014 (168 pp.).

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