The volume of HCOOH is 0.003 Lt and the volume of HCOONa is 0.017 LT.
Explanation:
For acidic buffer
PH = Pka + log [tex]\frac{salt}{acid}[/tex]
4.25 = 3.75 + log [tex]\frac{salt}{acid}[/tex]
log [tex]\frac{salt}{acid}[/tex] = 0.75
log [tex]\frac{salt}{acid}[/tex] = 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 = [tex]\frac{0.1*100}{0.5}[/tex] = 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,
[tex]\frac{v2*0.5}{100} /\frac{v1*0.5}{100}[/tex] = 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.
