Home
Class 11
CHEMISTRY
For the standardization of Ba(OH)2 solut...

For the standardization of `Ba(OH)_2` solution,"0.204g" of potassium acid phthalate was weighed which was then titrated with `Ba(OH)_2` solution.The titration indicated equivalence at "250ml" of `Ba(OH)_2` solution.The reaction involved is `KHC_8H_4O_4` + `Ba(OH)_(2)` → `H_(2)O` + `K^(+)` + `Ba^(+2)` + `C_(8)H_(4)O_(4)^(-2)` The molarity of the base solution is (K=39) options:- 1) 0.04M, 2) 0.03M 3) 0.02M,4) 0.01M

Promotional Banner

Similar Questions

Explore conceptually related problems

The pH of 0.05 M Ba (OH)_(2) solution is

The pH of 0.001 M Ba(OH)_(2) solution will be

A solution of Ba(OH)_(2) is standardized with potassium acid phthalate (abbreviated KHP), KHC_(8)H_(8)O_(4) (M=204). If 1.530 g of KHP is titrated with 34.50 mL of the Ba(OH)_(2) solution, what is the molarity of Ba(OH)_(2) ?

The volume of 3 M Ba(OH)^(2) solution required to neutralize completely 120 mL of 1.5M H_(3)PO_(4) solution is :

The pH of Ba(OH)_(2) solution is 13 . The number millimoles of Ba(OH)_(2) present in 10 ml of solution would be