Question

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. The resulting pH 4.35

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 52.0 mL ?

Answer 1

Ka = 5.61* 10^-6

Let x ml of 9*10^-3 mol of HA and 1*10^-3 mol of strong base make sol of pH 4.35

Now HA provides H+ ions and strong base provides us OH- ions

         x ml of HA provides x*9*10^-3 mol of H+ ions and y ml of strong base provides 10^-3 moles*y mloes of OH- ions

So remaining H+ ions = x*9*10^-3 - y*10^-3   = (9x-y) *10^-3

pH = - log (H+ ions) = 4.35

so H+ ions = 10^-4.35

so   ( 9x-y)* 10^-3   = 10^-4.35

so    9x-y = 10^-4.35/10^-3   = 10^ ^-1.35 =0.0447    ------(1)

     Also in II case x+ y =   52 ml            ---------(2)

Adding (1) and (2)

                     9x +1x = 52.0447

x =    5.20447 ml

Put this in (2)

5.20447 +y = 52

y = 52-5.20447 = 46.8 ml

Now H+ ions   in x ml = x * 9 *10^-3   =5.20447*9*10^-3 =0.04684

      OH - ions in y ml   = y *10^-3   = 46.8 * 10^-3   =0.0468

H+ ions - OH- ions = 0.04684-0.0468   = 4 * 10^-5

pH = - log H+ ions = - log (4*10^-5) = 4.398