Question

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)

Answer 1

Therefore
`pH` of the buffer solution is 6.96.

Step-by-step explanation:

The Henderson- Hasselbalch Equation:

`pH= pK_a+log ([salt])/([acid])`.

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

`=\frac{mass}{\textrm{molar mass}}`

`=(12)/(120)`

=0.1 mole

The number of mole of 8 gram of Na₂HPO₄ is

`=\frac{mass}{\textrm{molar mass}}`

`=(8)/(142)`

=0.056 mole.

Concentration of [H₂PO₄⁻ ] is

`=\frac{\textrm {Number of mole}}{Volume}`

`=(0.1)/(0.50)`

=0.2 M

Concentration of [HPO₄²⁻] is

`=\frac{\textrm {Number of mole}}{Volume}`

`=(0.056)/(0.50)`

=0.112 M

Therefore ,

`pH= pK_a+log ([salt])/([acid])`

`\Rightarrow pH=-log(6.2* 10^(-8))+log ((0.112)/(0.2))`

`\Rightarrow pH=6.96`

Therefore
`pH` of the buffer solution is 6.96.