In what volume ratio 0.1M HA( acid having ka=2*10E-5) is mixed with 0.2M NaOH to get a buffer of ph=5.

Question

In what volume ratio  0.1M HA( acid having ka=2*10E-5) is mixed with 0.2M NaOH to get a buffer of p^h=5.

Answer 1

The Henderson-Hasselbach equation:  pH = pK_a + log [A-]/[HA], which is just a variation of the equilibrium equation of Ka = [H+][A-]/[HA]

 

The K_a = 2 x 10^-5 M.  So we compute pKa = -log Ka:  pKa = -log 2 x 10-5 = -(-4.70) = 4.70  (I think it's acetic acid, by the way, which has a recognizable pKa).

 

Calculate [A-]/[HA] = 10^{(pH - pKa)} =10(5.00 - 4.70) = 10(0.3) = 2.0.  So the concentration of [A-] is twice that of [HA].

 

There is no [A-] solution available.  But if you add a strong base OH- to HA, you generate A-.  If you mix 1 vol 0.1 M HA and 1 vol 0.1 M NaOH, you get 2 vol 0.1 M A-.  If you mix 1 vol 0.1 M HA and 0.5 vol 0.2 M NaOH, you will get 1.5 vol A-.  But that is complete neutralization.  You want to add OH in a way to produce 2 A- with 1 HA remaining.

 

x HA + y OH --> (x - y) HA + (y) A-

 

y / (x - y) = 2  ==> y = 2x - 2y ==> 3y = 2x  ==>  y/x = 2/3.

 

This is a MASS ratio, NOT a volume ratio.  Since mass = concentration * volume, then y/x = (C1*V1)/(C2*V2) = 2/3. Since C1 = 0.2M and C2 = 0.1M, then 2/3=(0.2)V1/(0.1)V2  ==>  V1/V2 = 1/3.  In other words 1 vol of y (0.2 M NaOH) and 3 vol of x (0.M HA) produces the desired solution.

 

Verify.  The ratio is 1 vol OH-:3 vol HA.  So let's mix 100 ml 0.2 M NaOH and 300 ml 0.1 M HA.  That is (0.1 L)(0.2 mol/L) = 0.02 mol OH, and (0.3 L)(0.1 mol/L) = 0.03 mol HA.  This will create 0.02 mol A-, and leave 0.01 mol HA, in a total volume of 0.4 L.  The concentration will be (0.02 mol A-/0.4 L)/(0.01 mol HA/0.4 L), which is a concentration ratio of 2.  This will produce the desired pH 5.0.

 

ANSWER:  1 vol NaOH + 3 vol HA