Suppose that 9.4 mL of 3.6 M KOH(aq) is transferred to a 250 mL volumetric flask and diluted to the mark?

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2016-05-29 6:06 pm

It was found that 38.4 mL of this diluted solution solution was needed to reach the stoichiometric point in a titration of 9.4 mL of a phosphoric acid solution according to the reaction:
3 KOH(aq) + H3PO4(aq) →
K3PO4(aq) + 3 H2O(ℓ)
Calculate the molarity of the solution.
Answer in units of M.

An explanation would be awesome since my chemistry teacher does not teach and i want to be ready if he gives us a surprise quiz. Thanks!

回答 (1)

fooks

2016-05-29 6:24 pm

✔ 最佳答案

Firstly, consider the dilution of KOH solution to 250 mL :
M₁V₁ = M₂V₂
(3.6 M) × (9.4 mL) = M₂ × (250 mL)
Molarity of the diluted KOH solution, M₂ = 3.6 × (9.4/250) M = 0.1354 M

Secondly, consider the titration :
3 KOH(aq) + H₃PO₄(aq) → K₃PO₄(aq) + 3 H₂O(l)
Mole ratio KOH : H₃PO₄ = 3 : 1
No. of moles of KOH reacted = (0.1354 mol/L) × (38.4/1000 L) = 0.005199 mol
No. of moles of H₃PO₄ reacted = (0.005199 mol) × (1/3) = 0.001733 mol
Volume of H₃PO₄ solution used = 9.4 mL = 0.0094 L
Molarity of H₃PO₄ = (0.001733 mol) / (0.0094 L) = 0.184 M

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