CS3PO4 + Na2SO4 → Na3PO4 + Cs2SO4
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PH = pKa + log ![\frac{[base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bbase%5D%7D%7B%5BAcid%5D%7D%20)
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case