A 50 ml Sol of pH = 1 is mixed with 50 ml sol of pH =2 what will be the pH of mixture
Answers
Hello Dear.
Given conditions ⇒
In First Case,
Volume of the solution(V₁) = 50 mL.
pH = 1
∴ Concentration of H⁺ (M₁) = 10⁻¹
[∵ pH is less than 7, ∴ Solution is acidic.]
In Second Case,
Volume of the solution(V₂) = 50 mL.
pH = 2 [It is also acidic]
∴ Concentration of H⁺ (M₂) = 10⁻²
Now, Volume of the Final Solution(V) = V₁ + V₂
= 50 + 50
= 100 mL.
Since, both the solution above have the pH less than 7, i.e., both are acidic.
∴ MV = M₁V₁ + M₂V₂
⇒ M × 100 = 10⁻¹ × 50 + 10⁻² × 50
⇒ 100 M = 5 + 0.5
⇒ M = 5.5/100
⇒ M = 5.5 × 10⁻²
∴ Concentration of the H⁺ in the Final solution (or Mixture) is 5.5 × 10⁻².
Now, Using the Formula,
pH = -log [H⁺]
pH = -log[5.5/100]
pH = - log5.5 + log100
[∵ log(a/b) = loga - logb]
pH = log 100 - log 5.5
pH = 2 - log 5.5
[∵ log 100 = 2]
∴ pH = 2 - 0.74
[∵ log 5.5 = 0.74]
∴ pH = 1.26
Hence, the pH of the Final solution (or Mixture) is 1.26.
Hope it helps.
.
.
Mark as the brainliest.....