Question

Calculate the pH in the titration of 25.0 mL of 0.100M HCl with 0.200M NaOH. Next, calculate the pH after 6.25 mL of base have been added.

Answer 1

pH depends on the concentration of Acid

So, before titration

HCl as strong acid, it dissociates completely

pH = - log [H⁺] = - log [0.100] = - log [10^-1] = 1 (Using log 10^m = m )

Now, after adding 6.25 mL of 0.200M NaOH

HCl + NaOH ------> NaCl + H₂O

The reaction goes by the stoichiometry of 1: 1

So, the moles of HCl after the addition of NaOH = (25 - 12.5 ) * 10^-4 = 12.5 * 10^-4 (Conc. = mole / Volume )

Now, Conc of HCl in the solution = ( 12.5 * 10^-4 ) / (25 + 6.25 mL) = ( 12.5 * 10^-4 ) / (31.25 * 10^-3 L) = 0.4 * 10 ^-1

Now pH = - log [ 0.4 * 10 ^-1 ] = - log [ 4 * 10^-2 ] = - { log 4 + log 10^-2 } = - { 0.602 - 2 } = - { - 1.398 } = 1.398