Write the equations for the stepwise dissociation
ofpyrophosphoric acid, h4p2)7. identify all conjugate
acid-basepairs.

    H₄P₂O₇    +    H₂O   ⇄                 H₃P₂O₇⁻            +             H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid

  H₃P₂O₇⁻       +  H₂O   ⇄                H₂P₂O₇⁻²            +            H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid  

     H₂P₂O₇⁻²  +  H₂O     ⇄                    HP₂O₇⁻³         +         H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid

    HP₂O₇⁻³      +  H₂O   ⇄                     P₂O₇⁻⁴           +           H₃O⁺

Weak acid         Base         Strong conjugate base         conjugate acid

Explanation:

Pyrophosphoric acid → H₄P₂O₇

It's a poliprotic acid, which is able to release more than two protons. Pyrophosphoric acid is generated when we add 2 water to phosphorus (V) oxide .

P₂O₅ + 2H₂O  →  H₄P₂O₇

As it is able to release 4 protons, it will have 4 Ka for each equilibrium.

    H₄P₂O₇    +    H₂O   ⇄                 H₃P₂O₇⁻            +             H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid

  H₃P₂O₇⁻       +  H₂O   ⇄                H₂P₂O₇⁻²            +            H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid  

     H₂P₂O₇⁻²  +  H₂O     ⇄                    HP₂O₇⁻³         +         H₃O⁺

Weak acid         Base         Strong conjugate base        conjugate acid

    HP₂O₇⁻³      +  H₂O  ⇄                     P₂O₇⁻⁴           +           H₃O⁺

Weak acid         Base         Strong conjugate base         conjugate acid

As the water, is always receiving one H⁺ behaves as a base, so the H₃O⁺ will be the conjugate acid

H₄P₂O₇ is a weak acid, it donates the H⁺ to water, so the anions will be the strong conjugate base. They will react in water donating the protons until become P₂O₇⁻⁴

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml