Answer:
The pH of the resultant solution is 2.4
Explanation:
From the data
V_total=V_w+V_a
V_total=400+200 ml
Now the concentration of solution is given as
[tex]M_1V_1=M_2V_2\\M_2=\frac{M_1V_1}{V_2}\\M_2=\frac{0.30 \times 0.2}{0.6}\\M_2=0.1 M\\[/tex]
The reaction equation is given as
Equation [tex]HCOOH + H_2O \rightarrow HCOO^- +H_3O^+\\[/tex]
Initial Concentration is 0.1 0 0
Change is -x x x
_______________________________________________
At Equilibrium 0.1-x x x
Now the equation for K_a is given as
[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}[/tex]
Here Ka, for pKa-=3.75, is given as
[tex]K_a=10^{-pKa}\\K_a=10^{-3.75}\\K_a=1.78 \times 10^{-4}[/tex]
Substituting this and concentrations at equilibrium in equation of Ka gives
[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}\\1.78 \times 10^{-4}=\frac{xx}{0.1-x}\\[/tex]
Solving for x
[tex]1.78 \times 10^{-4} (0.1-x)=x^2\\[/tex]
Here 0.1-x≈0.1 so
[tex]1.78 \times 10^{-5} =x^2\\x=\sqrt{1.78 \times 10^{-5}}\\x=4.219 \times 10^{-3}[/tex]
As [tex][H_3O^+]=[H^+][/tex]=x so its value is 4.219 x10^(-3) so pH is given as
[tex]pH=-log[H^+]\\pH=-log (4.219 \times 10^{-3})\\pH=2.37 \approx 2.4[/tex]
So the pH of the resultant solution is 2.4
Answer:
2.4
Explanation:
Given that:
Initial Concentration (M₁) of our formic acid from the question= 0.30 M
Initial volume (V₁) of the formic acid = 200 mL
Volume of water added = 400 mL
pKa= 3.75
∴ we can determine the total volume of the final solution by the addition of (Initial volume (V₁) of the formic acid) + (Volume of water added)
= 200 mL + 400 mL
= 600 mL
We can find the final concentration using the formula;
M₁V₁ = M₂V₂
M₂= [tex]\frac{0.30*200}{600}[/tex]
M₂= 0.1M
∴ The concentration of the diluted formic acid(since water is added to the initial volume) = 0.1M
From our pKa which is = 3.75
Ka of formic acid (HCOOH) = [tex]10^{-3.75}[/tex]
= 1.78 × 10⁻⁴
If water is added to the formic acid; the equation for the reaction can be represented as;
HCOOH + H₂O ⇒ HCOO⁻ + H₃O
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x x x
Ka = [tex]\frac{x^2}{0.1-x}[/tex] = 1.78 × 10⁻⁴
x² = (0.1 × 1.78 × 10⁻⁴) since x = 0
x² = 1.78 × 10⁻⁵
x = [tex]\sqrt{1.78*10^{-5}}[/tex]
x = 0.004217
x = [tex][H_3O^+]=[H^+][/tex] = 0.004217 M
pH = [tex]-log[H^+][/tex]
= -log (0.004217)
= 2.375
≅ 2.4 (to one decimal place)
We can thereby conclude that the final pH of the formic acid solution = 2.4
