Calculate the volume of 0.500 M HCOOH and 0.500 M HCOONa required to prepare 0.100 L of pH = 4.25 buffer with a buffer
strength of 0.100 M. The pk, of HCOOH is 3.75.
нсоон:
HCOONa:

The volume of HCOOH is 0.003 Lt and the volume of HCOONa is 0.017 LT.

Explanation:

For acidic buffer

PH = Pka + log

4.25 = 3.75 + log

log = 0.75

log = 5.62 is considered as equation 1.

Let V₁ ml of HCOOH and V₂ml of HCOONA was mixed and the final volume is 100 ml .

V₁ *0.5 + V₂ *0.5 =0.1 * 100

V₁+V₂ = = 20 is the second equation.

Now concentration of salt and acid are

V₂ [salt] = 0.5/100  and v₁ [acid] = 0.5/100

by substituting these in equation 1 we get,

= 5.62

V₂ = V₁ * 5.62 is the 3 rd equation

by solving 2 and 3 we get,

5.62 V₁ + V₁ = 20

= 6.62 V₁ = 20

V₁ = 3.02 ml

V₁ = 0.003 LT

V₂ = V₁ * 5.62

V₂ = 0.017 LT

Hence the volume of  HCOOH is 0.003 Lt and the volume of HCOONa is 0.017 LT.

am i too late?

explanation: