In this experiment, an acid and a base reagent will be prepared and standardize for use in experiments #7 and #9.
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
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
= 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
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
Calculation of concentration for the base (NaOH):
... to the moles of KHP titrated. n (mol) n (mol) NaOH KHP .Before you titrate a KHP sample, begin ... determine how much acid you will be titrating in Trial # 1. Note: There are only two samples of ... the difference between the two mass measurements (ex. if Trial # 1 gives a mass of 240 g/mol and Tiral ... with a practice titration of 25 mL of a 0.1 M HCl ...
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