6. What is the pH of the buffer that results when 12.0 g of NaH2PO4 and 8.00 g of Na2HPO4 are diluted with water to a volume of 0.50 L? (Ka of H2PO4– = 6.2 x10–8, the molar masses of NaH2PO4 and Na2HPO4 are 120.0 g/mol and 142.0 mol, respectively)

Therefore of the buffer solution is 6.96.

Explanation:

The Henderson- Hasselbalch Equation:

.

The acid will be H₂PO₄⁻ and the salt of the acid will be HPO₄²⁻.

Given the molar mass of  NaH₂PO₄ is 120.0 g/mol and the molar mass of Na₂HPO₄ is 142.0g/mol.

The number of mole of 12 gram of NaH₂PO₄ is

=0.1 mole

The number of mole of 8 gram of Na₂HPO₄ is

=0.056 mole.

Concentration of [H₂PO₄⁻ ] is

=0.2 M

Concentration of [HPO₄²⁻] is

=0.112 M

Therefore ,

Therefore of the buffer solution is 6.96.

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