Calculate the pH of a buffer solution obtained by dissolving 25.025.0 g of KH2PO4(s)KH2PO4(s) and 38.038.0 g of Na2HPO4(s)Na2HPO4(s) in water and then diluting to 1.00 L.

pH = 7.37

Explanation:

The buffer H₂PO₄⁻/HPO₄²⁻ has as pKa 7.21. To find the pH of a buffer you can use H-H equation:

pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

Where [HPO₄²⁻] and [H₂PO₄⁻] are molar concentrations of each species. As volume is 1.00L, [HPO₄²⁻] and [H₂PO₄⁻] are MOLES.

Moles of 25.0g of KH₂PO₄ (Molar mass: 136.086g/mol):

25.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.1837 moles KH₂PO₄ = moles H₂PO₄⁻

Moles of 38.0g of Na₂HPO₄ (Molar mass: 141.96g/mol):

38.0g KH₂PO₄ ₓ (1mol / 141.96g) = 0.2677 moles Na₂HPO₄ = moles HPO₄²⁻

Replacing in H-H equation:

pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

pH = 7.21 + log₁₀ [0.2677] / [0.1837]

pH = 7.37

Ithink chemical bonding

Using dilution equationm1v1 =m2v2 ; where m1 and m1 are the intiial and final concentration, and v1 and v2 are the initial and final volume.5  × 5 m = 8 l  × m2m2 = 25/8 = 3.125 mtherefore; the resulting molarity is 3.125 m