Question

QUESTION 4 Use the sample results below to complete the Report Form Laboratory Temperature 22.0C Concentration of Stock Fe(NO) soin. 0.00200 M Concentration of Stock KSCN soin. 0.00200 M Trial Fe(NO3h (mL) KSCN (ml) H2O (ml) Total Volume (ml) % Transmitance 1 5.00 5.00 0.000 10.0 28.0 2 5.00 4.00 1.00 10.0 32.8 3 5.00 3.00 2.00 10.0 41.2 4 5.00 3.00 10.0 51.0 200 Calculate the equilibrium constant (Kec) of each trial as well as the average Keq (Ave. Keel Please fill in each blank with integers, e.g. 100, 200, 300 etc. and round answers to the correct number of significant figures. Trial Kog 1 2 3 4 Ave. Kog

Answer 1

Trial FeWo 3 KSCN (m) (40) m) Total volume of Trans the (ML) (mL) 5.00 500 0.0 10.0 28.0 No. 5.00 4.00 1.0 10.0 32.8 3 5.00 300 2.0 10.0 41.2 4 5.00 2.00 13.0 10.0 510 Absorbence (fesor) IM) 000010 0.553 000008 0.484 0.00006 0.385 0.00009 0.292 Reaction F e +3 + SCN (0.002m] → FeSON (0.002 M Calculation > Absorbance = 2- log (% transmitenee) Abs 2 - log (28) = 0.553 Trial 1.

Goodluck Page No. Trial 2. Dale Abs = 0.484 2-log (328) Trial 3 Abs= 2-log (412) 0. 385 Total 4. Absə 2-log (510) = 0.2.92 Concentration of FeS EN + 2 CKSCNJ [FescN+2 cm, v.) (M₂ V₂2 0.0002) (5) = Me (10) 0.0001 Trial. a (0.002) (4) 000008 = M2 (10) Trial 3. (0.002) (3) = M2 (10) 0.00006 Trial. 4. Co.002) (2) = M₂ (10) 0,00004) Part-I-> Calibration Curve From graph , slope 4500=m Intercepta. 0.09 7 = mx + c. Abs = (slupe) (resent & Intercept

GoodLuck Page No. Date Tae - T- F e +3 + 0.002 M SON 0.002 M. PESCNT Trial 5 m2 (0.002) (6) = M. (10) 0.001 5m2. (0.002) (s) = M₂. (0) -0.001 E 0.001-X 0.001-X X (FeSON + Abs Intocept = 0.553-0.09 Slope 4500 At equilibrium » [Fe sont j = x=0.00010 M [Fe +3] = 0.0001 -0.00010 = 0.0009 m [SCNJ = 0.001 - 0.00 0 10 = 0.0oog M keq = [rescut [fes] [soul [keq = 123 0.oool 0.0009) 0.0005) Trial. 2. SmL (0.0025 M2 (10) 4mL. (0,004) (4)-M: (10) I с ono 01 -X 0.0008 -X tx E 0.001-X 0.0008-X X.