Question

A.Calculate the pH of a buffer that is 0.180 M in NaHCO3 and 0.350 M in Na2CO3.

B.Calculate the pH of a solution formed by mixing 65 mL of 0.18 M NaHCO3 with 75 mL of 0.28 M Na2CO3.

Answer 1

A)

This is a buffer so

pH = pKa + log(base/Acid)

pKa = 10.32 ( from data bases)

the acid = NaHCO3; since it will form; Na+ and HCO3- in solution, the HCO3- donates its H+ ion to form CO3-2 ions

the base = Na2CO3; since it will form; 2Na+ and CO3-2 in solution, the CO3-2 will accept the H+ ion, that is a base

then

pH = 10.32 + log(0.35/0.18) = 10.6087

pH = 10.61

B)

concnetrations will change, then

Total Voluem = V1+V2 = 65+75 = 140 ml

mol of NaHCO3 = MV = 65*0.18 = 11.7mmol

mol of Na2CO3 = MV = 75*0.28 = 21 mmol

then

[NaHCO3 ] = 11.7/140 = 0.083571 M

[Na2CO3 ] = 21/140 = 0.15M

then; similar to A

pH = pKa + log(base/Acid)

pKa = 10.32 ( from data bases)

pH = 10.32 + log(0.15/0.083571)

ph = 10.574