Respuesta :

The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)

1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
                                             = 0.0563 moles
 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
     = 7.218

pH is the estimation of the acidity and the basicity of the solution. The pH of the soft drink with buffer ingredients will be 7.218.

What is pH?

pH is the negative logarithm of the hydrogen ion concentration in an aqueous solution.

Moles of sodium dihydrogen phosphate is calculated as:

[tex]\begin{aligned}\rm moles &= \dfrac{6.6}{120} \\\\&= 0.055 \;\rm moles \end{aligned}[/tex]

The molar concentration of dihydrogen phosphate is calculated as,

[tex]\begin{aligned} &= \dfrac{\text{Moles of dihydrogen phosphate}}{\text {Volume }}\\\\&= \dfrac{0.055}{355 \times 10^{-3}}\\\\&= 0.155 \;\rm M\end{aligned}[/tex]

The complete dissociation of monosodium phosphate can be shown as,

[tex]\rm Na_[2}HPO_{4} (aq) \rightarrow 2Na^{+} (aq) + HPO_{4}^{2-} (aq)[/tex]

When 1 mole of monosodium phosphate = 142 g/mol

Then, moles =  0.0563 moles

The molar concentration of dihydrogen phosphate will be:

[tex]\begin{aligned} &= \dfrac{0.0563}{( 355 \times 10^{-3})}\\\\&= 0.1586\;\rm M\end{aligned}[/tex]

The partial dissociation of both the weak acid is given as:

[tex]\begin{aligned}\rm [H^{+}] &= \rm Ka \times \dfrac{([H_{2}PO_{4}^{-}]}{[HPO_{4}^{2-}]}\\\\ &= (6.20 \times 10^{-8}) \times (\dfrac{0.155}{0.1586})\\\\ &= 6.059 \times 10^{-8}\;\rm M \end{aligned}[/tex]

pH can be calculated as:

[tex]\begin{aligned} \rm pH &= \rm - log[H^{+}]\\\\ &= \rm - log (6.059 \times 10^ {-8})\\\\ & = 7.218\end{aligned}[/tex]

Therefore, 7.218 is the pH.

Learn more about pH here:

https://brainly.com/question/9041846

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