During a titration experiment, when you add 52.00mL of 0.98 M Ba(OH)2 to 42.00 mL of...

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During a titration experiment, when you add 52.00mL of 0.98 M Ba(OH)2 to 42.00 mL of...

During a titration experiment, when you add 52.00mL of 0.98 M Ba(OH)₂ to 42.00 mL of H₃A to reach the equivalence point. What is the concentration of H₃A? The MW of H₃A = 69.0 g/mol

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Answer 1

Answer #1

Ans :

Balanced reaction is given as :

3Ba(OH)₂ + 2H3A = Ba₃A₂ + 6H₂O

Using the relation : M1V1 / n1 = M2V2 / n2

here M1 = 0.98 M

V1 = 52.00 mL

n1 = 3 mol

V2 = 42.00 mL

n2 = 2 mol

putting values :

(0.98 x 52.00 ) / 3 = ( M2 x 42.00 ) / 2

M2 = 0.81 M

so the concentration of H3A will be 0.81 M.

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Answer 2

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