Question

Calculate the pH of a solution by mixing 3.29 g of "bicarb", NaHCO3, and 4.15 g of NA2CO3 in water to form 4.71 L of solution. The Ka1 for H2CO3 IS 4.31X10^-7 AND THE KA2 IS 4.40X10^-11.

Answer 1

NaHCO3 moles = mass / Molar mass of NaHCO3

               = ( 3.29 g) / ( 84 g/mol) = 0.03916 mol

Na2CO3 moles = mass / Molar mass of Na2CO3

                        = ( 4.15g) / ( 106 g/mol) = 0.03915 mol

we have pH of buffer given by formula

pH = pka + log ( conjugate base moles) / ( acid moles)

where acid is NaHCO3 and its conjugate base is Na2CO3 ,

pka of HCO3-   (or NaHCO3) = -log Ka2 = -log ( 4.4 x 10^-11) = 10.36

pH = 10.36 + log ( 0.03915 /0.03916)

=   10.36