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17.131) A 24.0 −mL volume of a sodium hydroxide solution requires 19.5 mL of a 0.193 M hydrochloric acid for neutralization. A 11.0 −mL volume of a phosphoric acid solution requires 34.8 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution

Respuesta :

Answer:

The concentration of Phosphoric acid required for the neutralization described = 0.165 M

Explanation:

Given,

Volume of NaOH = V = 24.0 mL

Concentration of HCl = Cₐ = 0.193 M

Volume of HCl = Vₐ = 19.5 mL

NaOH + HCl -----> NaCl + H₂O

1 mole of NaOH reacts with 1 mole of HCl

Using the equivalence point expression

(CₐVₐ)/(CV) = (nₐ/n)

where

Cₐ = concentration of acid = 0.193 M

Vₐ = volume of acid = 19.5 mL

C = concentration of base = ?

V = volume of base = 24.0 mL

nₐ = Stoichiometric coefficient of acid in the balanced equation = 1

n = Stoichiometric coefficient of base in the balanced equation = 1

(CₐVₐ)/(CV) = (nₐ/n)

(0.193 × 19.5)/(C × 24) = 1

(C × 24) = 3.7635

C = (3.7635/24) = 0.157 M

This NaOH is then reacted with phosphoric acid.

Phosphoric acid = H₃PO₄

3NaOH + H₃PO₄ --------> Na₃PO₄ + 3H₂O

3 moles of NaOH reacts with 1 mole of Phosphoric acid.

Using the equivalence point expression

(CₐVₐ)/(CV) = (nₐ/n)

where

Cₐ = concentration of acid = ?

Vₐ = volume of acid = 34.8 mL

C = concentration of base = 0.157 M

V = volume of base = 11.0 mL

nₐ = Stoichiometric coefficient of acid in the balanced equation = 1

n = Stoichiometric coefficient of base in the balanced equation = 3

(CₐVₐ)/(CV) = (nₐ/n)

(Cₐ × 11)/(0.157 × 34.8) = (1/3)

Cₐ × 11 × 3 = 0.157 × 34.8 × 1

33Cₐ = 5.457

Cₐ = (5.457/32)

Cₐ = 0.165 M

Hence, the concentration of Phosphoric acid required for the neutralization described = 0.165 M

Hope this Helps!!!

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