Question

Calculate the ph of a solution prepared by mixing 15.0ml of 0.10M NaOH and 30.0ml of 0.10M benzoic acid solution. (Benzoic acid is monoprotic, it's dissociation constant is 6.5×10^-5)

Answer 1

Benoaic acid ( C6G5COOH) moles = M x V = 0.1 x ( 30/1000) = 0.003

NaOH moles = M x V = 0.1 x 15/1000 = 0.0015 = OH- moles

we have reaction   C6H5COOH + OH -(aq) <----> C6H5COO-(aq) + H2O (l)

after reactiong with OH- , C6H5COOH moles = 0.003-0.0015 = 0.0015

moles of C6H5CO- = OH- moels reacted = 0.0015 , total solution vol = 15+30 = 45 ml = 0.045 L

now pH = pka + log [C6H5COO-]/[C6H5COOH]   where pka =-log Ka = -log ( 6.5x10^ -5) = 4.178

pH = 4.187 + log ( 0.0015/0.045) / ( 0.0015/0.045)

= 4.187