MOHR Method (K2CrO4 indicator) EXP.NO.4 Precipitation A/ Preparation of standard 0.1N NaCl soln. & standardization of AgNO3 soln. B/ Determination of [Cl-] in unknown solution. MOHR Method (K2CrO4 indicator)

At the end point Mohr method depend on (differential precipitation), so as shown at the E.P the reddish brown ppt. Will from with the indicator which is (Ag2CrO4) which more soluble than the precipitates (AgCl & AgBr),so the precipitate (Ag2CrO4) will form after complete precipitation of all (Cl ) as AgCl or (Br) as AgBr . Indicator K2CrO4:- How much we need to indicate precipitation of Ag2CrO4 at the E.P? at the E.P the only (Ag) present is from the solubility of AgCl PPt. which is equal to (1.3 × 10-5 ) so [Ag+]2[CrO42-]= Ksp (Ag2CrO4) to start the ppn. [1.3×10-5]2[CrO42-]=1.1×10-12 [CrO42-]=(1.1×10-12)/1.7×10-10=10-2? 0.01M

There is a systematic error (E) which came from the precipitation of the (reddish brown) ppt. of (Ag2CrO4) at the E.P ,so we must determine volume of titrant (AgNO3) which required for this precipitation which determined by BLANK titration, where A= acceptable value E= error (systematic error) O= observed value (VAgNO3 needed for sample) Blank titration carried out by taking volume of H2O (which is equal to the volume of analyte soln. at E.P.) and some amount of white ppt. (0.2 gm CaCO3) and the same amount of indicator in a conical flask and titrating it with AgNO3 soln., and from this result we find the volume of AgNO3 needed to form the reddish brown ppt. Ag2CrO4 which is equal to (E).

It must be mentioned that the titration should be carried out in neutral medium or in a very faint alkaline soln. (with pH range 6.5-9), because in acid soln. the following reaction occurs: so the amount of indicator CrO4 will decrease in the soln. and the ionic product [Ag+][CrO42-] may be exceed Ksp of AgCrO4, so the ppt. will not form at E.P. in alkaline soln. silver hydroxide (Ksp= 2.3×10-10) may be precipitated. So the alkaline soln. the HAc acid added followed by small amount of CaCO3 and for acid soln. a small amount of NaHCO3 or CaCO3 added to adjust the pH. Titration of iodide [I-] or [SCN-] is not successful because (AgI) or (AgSCN) absorb [CrO42-] ion so strongly and indistinct E.P. is obtained.

4-Determine (E) by doing blank titration as follow: Procedure: A/ Preparation of standard 0.1N NaCl soln. & standardization of AgNO3 soln. 1-Weigh out accurately (0.5844 gm) of solid pure NaCl on a watch glass by a sensitive balance, dissolve it in D.W., transfer it quantitatively to a (100ml vol. flask), complete the volume to the mark. 2-Pipette (10ml) of the prepared soln. to a conical flask, add (0.5ml) of (K2CrO4) indicator. 3-Titrate with AgNO3 soln. form burette until a permanent faint reddish-brown ppt. will form, record V AgNO3 needed as (O) value. 4-Determine (E) by doing blank titration as follow: {pipette (20ml) D.W. to a conical flask, add (0.2gm) of solid CaCO3, add (0.5ml) of K2CrO4 indicator, titrate it with (10 time diluted AgNO3) soln. from a micro-burette. Record volume of AgNO3 needed after it, divided it by 10} record volume as (E), A= O-E 5- Calculate N of AgNO3 soln. NAgNO3×(A)= NNaCl×VNaCl

B/ Determination of [Cl-] in unknown sol. 1- Pipette (10ml) of unknown [Cl-] soln.to a conical flask ,add (0.5ml) of K2CrO4 indicator . 2- Titrate with the same (AgNO3) soln. from burette until (faint reddish - brown) ppt form. Record VAgNO3 as (O) also [A=O - E] NAgNO3 × (A) = NCl- × VCl- Calculate (gm/L)[Cl-] present in the sample . NCl- × 35.5 = (gm/L)Cl-