A500.0 ml buffer solution is 0.10 m benzoic acid and 0.10 m sodium benzoate has an initial ph of 4.19. what is the ph of the buffersolution upon addition of 0.010 mol ofnaoh? the kafor benzoic acid is 6.5 • 10-5

New pH = 4.27

Adding 0.01 mol of NaOH would increase the amount (in mol, 0.10) of sodium benzoate and decrease the amount (in mol, 0.10) of benzoic acid. This is because NaOH would react with benzoic acid to form more sodium benzoate. NaOH would get used up in the solution. Adding (sodium benzoate + NaOH) and subtracting (benzoic acid + NaOH) would give you 0.11 mol of sodium benzoate and 0.09 mol of benzoic acid. New pH = pKa + log [mol of sodium benzoate/mol of benzoic acid]. Hope this helps!