Standardization of sodium hydroxide against potassium hydrogen pthalate?

does anyone know how to find the concentration Sodium hydroxide (.10 M) titrated with .700 grams of KHC8H4O4? three trials were titrated to 38.4,37.0 and 36.1 mL

Best Answer

1. Determine the molar mass of KHC8H4O4 = 204.1 g/mol

2. Determine the number of moles:
(0.700g)/(204.1g/mol) = 0.00343 mol

3. Determine the molarity of the three trials. This will show you the concentration of NaOH.
a. 0.00343 mol/ 0.0384L = 0.089 M
b. 0.00343 mol/ 0.0370L = 0.093 M
c. 0.00343 mol/ 0.0361L = 0.095 M

4. Get the average of the concentration of the three trials:
( 0.089 M + 0.093 M + 0.095 M)/3 = 0.092 M

Hence, the actual concentration of your 0.1M NaOH is 0.092M.

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