Find the Volume of NaOH?

Question

What volume of 0.100M NaOH is needed to make 100.0mL of a pH6.00 buffer if one starts with 50.0mL of 0.100M KHP? Ka2 = 3.1 x 10-6.

Answer 1

Use the Henderson-Haselbach equation

pH= pKa2+ log[P(-2)]/[HP(-)] =>
[P(-2)] / [HP(-)]= 10^(pH-pKa2) =10^pH*10^-pKa2=Ka2*10^pH =
= (3.1*10^-6)*10^6 =3.1 => [HP(-)] = [P(-2)]/3.1=(1)

But [HP(-)]total= [P(-2)]+[HP(-)] (2) and
[HP(-)]total =M*V (3)

So (2) is written with the help of (1),(3) as
M*V = [P(-2)]+[P(-2)]/3.1 =>
M*V= (4.1/3.1) [P(-2)] =>
[P(-2)]= (3.1/4.1)* M*V =(3.1/4.1)*0.1*0.050 = 3.78*10^-3 M

We can assume that all of this comes from the reaction of HP(-) with NaOH. Thus
mole P(-2) =mole NaOH =>
M1*Vtotal = M2*Vx =>
0.00378*(50 + Vx) =0.1Vx =>
Vx= 0.00378*50/(0.1-0.0038) = 1.96 =2 mL

Note that in the last equation I didn't convert mL into L since the convertion factor is simplified.

If you don't want to make any assumptions and have a more accurate solution (though I doubt your teacher is looking for the more accurate solution since it is more complicated)
Then mole NaOH= M*V= 0.1Vx
mole HP(-)= mole KHP= 0.1*0.05 =0.005
Because of the reaction you have forming
mole P(-2)= mole NaOH = 0.1Vx and remaining
mole HP(-) = 0.005-0.1Vx

so [P(-2)]o= 0.1Vx/(0.05+Vx) (4) and
[HP(-)]o = (0.005-0.1Vx) / (0.05+Vx) (5)

.. .. .. .. .. .. .. HP(-) <=> H+ + P(-2)
Initial .. .. .. [HP(-)]o .. .. .. .. .. [P(-2)]o
Dissociate .. . y
Produce .. .. .. .. .. .. .. .. y .. .. .. y
At Equil. .. [HP(-)]o-y .. ..y .. .. [P(-2)]o+y

Ka= [H+][P(-2)]/ [HP(-)] = y*([P(-2)]o+y)/ ([HP(-)]o-y)
Substitute using (4), (5) and the fact that y=[H+]= 10^-pH= 10^-6
and solve the equation for Vx.

Answer 2

dear dont disturb me

take 6.0 pH buffer tab and dissolve it in required quantity of water