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"No tommarrow without Today" - Asra

Chem Lab

  • Date Submitted: 03/14/2010 12:13 PM
  • Flesch-Kincaid Score: 101.7 
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PURPOSE

In this experiment, an acid and a base reagent will be prepared and standardize for use in experiments #7 and #9.

RESULTS
STANDARDIZATION OF NaOH
KHP (g) NaOH (mL)
TRIAL 1 0.4239 22.56 * 3 drops of Phenolphthalein were added.
TRIAL 2 0.4095 21.08 In both trials.
TRIAL 3 0.4235 20.76
AVG 0.4190 21.47

STANDARDIZATION OF HCl
NaOH (mL) HCl (mL)
TRIAL 1 25 25.01 * 3 drops of Methyl Purple were added.
TRIAL 2 25 24.52 in both trials.
TRIAL 3 25 24.10
AVG 25 24.54

 CALCULATIONS:

1.0L and 0.1N Preparation:

Mass in g   = M   x   mm   x   V(L)

0.1M   x   40g   x   1L   =   4.0g of NaOH

From bottle = 36.38% ≈ 37% and 1.185 g/mol density.

Assuming that 1L of solution is prepared:

Mass of HCl     = 1.185 g/mol     X   36 g
  100 g

= 426.6 g HCl

Molarity     = # of moles →     g solute → 426.6 g
(mm) (V(L)) (mm) (V(L)) (36.5g/mol)(1000mL)

                  = 11.69M   in class we used 11.96M


Now:
C1V1 = C2V2   where   = C1   →   12M  
C2   →   0.1M  
V2   →   1.0L  
V1   →   ?

then V1 =   C2V2     → (0.1M)(1.0L) → 8.55mL   ≈   in class we used 8.36 mL  
      C1 12M

CONCLUSION
Calculation of concentration for the base (NaOH):
 Average of titrant (NaOH):
Mass KHP(g) → 0.4190g =   0.09556M ≈ 0.0956M
(MW KHP(g/mol))(Vol NaOH(L)) (204.22g/mol)(0.02147L)

 Average of titrant (HCl):
(NNaOH)(VNaOH) → (0.0956M)(0.025mL) =   0.0974M
VHCl 0.02454L

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