Question

If 200 ml of 0.30 M formic acid is added to 400 ml of water, what is the resulting pH of the solution? Round the answer to one decimal place. pKa= 3.75

Answer 1

Answer:

The pH of the resultant solution is 2.4

Explanation:

From the data

• [HCOOH]_0 is 0.30
• V_w is 400 ml
• V_a is 200 ml
• V_total is given as

          V_total=V_w+V_a

          V_total=400+200 ml

• V_total=600ml

Now the concentration of solution is given as

$M_1V_1=M_2V_2\\M_2=\frac{M_1V_1}{V_2}\\M_2=\frac{0.30 \times 0.2}{0.6}\\M_2=0.1 M$

The reaction equation is given as

Equation                            $HCOOH + H_2O \rightarrow HCOO^- +H_3O^+$

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

$K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}$

Here Ka, for pKa-=3.75,  is given as

                   $K_a=10^{-pKa}\\K_a=10^{-3.75}\\K_a=1.78 \times 10^{-4}$

Substituting this and concentrations at equilibrium in equation of Ka gives

$K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}\\1.78 \times 10^{-4}=\frac{xx}{0.1-x}$

Solving for x

$1.78 \times 10^{-4} (0.1-x)=x^2$

Here 0.1-x≈0.1 so

$1.78 \times 10^{-5} =x^2\\x=\sqrt{1.78 \times 10^{-5}}\\x=4.219 \times 10^{-3}$

As $[H_3O^+]=[H^+]$=x so its value is 4.219 x10^(-3) so pH is given as

$pH=-log[H^+]\\pH=-log (4.219 \times 10^{-3})\\pH=2.37 \approx 2.4$

So the pH of the resultant solution is 2.4

Answer 2

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= $\frac{0.30*200}{600}$

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) = $10^{-3.75}$

                                             =  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 = $\frac{x^2}{0.1-x}$ = 1.78 × 10⁻⁴

x² = (0.1 × 1.78 × 10⁻⁴)      since x = 0

x² =  1.78 × 10⁻⁵

x = $\sqrt{1.78*10^{-5}}$

x = 0.004217

x = $[H_3O^+]=[H^+]$ =  0.004217 M

pH = $-log[H^+]$

= -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