Question

Determine the volume (in mL) of 0.442 M potassium hydroxide (KOH) that must be added to 475 mL of 0.0992 M ascorbic acid (C₅H₇O₄COOH) to yield a pH of 3.89.



Assume the 5% approximation is valid and report your answer to 3 significant figures. A table of pKa values can be found here.

Answer 1

Now we know here the reaction of acid + base gives salt and water

But here the acid is weak and base is strong.

Lets see the reaction

Ascorbic Acid + NaOH -------------> Ascorbic acid salt + H20

Initial 0.475*0.0992 x*0.442 0 0

Final   0.475*0.0992-x*0.442   0 x*0.442

[Ascorbic acid at eq.] = 0.475*0.0992-x*0.442 / (0.475+x)

[Salt of ascorbic acid] = x*0.442/(0.475+x)

Here we see that Ascorbic acid and its salt will form a buffer solution . So now we need to know what should be amount of NaOH added to have the buffer as pH =3.89

We know that pH of buffer solution is,

pH = pKa + log( [salt]/[acid] )

we know pKa of ascorbic acid = 4.2

So we have

3.89 = 4.2 + log [salt]/[acid]

0.31 = log [acid]/[salt]

2.04 = [acid]/[salt] .......................1

Now using concentrations of ascorbic acid and salt at equilibrium we get ,

2.04 = (0.475*0.0992-x*0.442 / (0.475+x) ) / (x*0.442/(0.475+x))

2.04 = 0.475*0.0992-x*0.442 / x*0.442

0.9 x = 0.0475 - 0.442x

1.342 x = 0.0475

x =0.0475/1.342 = 0.035 L = 35 mL

So amount of NaOH added = 35 mL