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

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!!!

answer:

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