Respuesta :
Answer:
Therefore [tex]pH[/tex] of the buffer solution is 6.96.
Explanation:
The Henderson- Hasselbalch Equation:
[tex]pH= pK_a+log \frac{[salt]}{[acid]}[/tex].
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
[tex]=\frac{mass}{\textrm{molar mass}}[/tex]
[tex]=\frac{12}{120}[/tex]
=0.1 mole
The number of mole of 8 gram of Na₂HPO₄ is
[tex]=\frac{mass}{\textrm{molar mass}}[/tex]
[tex]=\frac{8}{142}[/tex]
=0.056 mole.
Concentration of [H₂PO₄⁻ ] is
[tex]=\frac{\textrm {Number of mole}}{Volume}[/tex]
[tex]=\frac{0.1}{0.50}[/tex]
=0.2 M
Concentration of [HPO₄²⁻] is
[tex]=\frac{\textrm {Number of mole}}{Volume}[/tex]
[tex]=\frac{0.056}{0.50}[/tex]
=0.112 M
Therefore ,
[tex]pH= pK_a+log \frac{[salt]}{[acid]}[/tex]
[tex]\Rightarrow pH=-log(6.2\times 10^{-8})+log (\frac{0.112}{0.2})[/tex]
[tex]\Rightarrow pH=6.96[/tex]
Therefore [tex]pH[/tex] of the buffer solution is 6.96.
