Question

Formic acid has a Ka of 1.77*10-4. To 55.0 mL of 0.25 M solution 75.0 of 0.12 M NaOH is added . What is the resulting pH.

Answer 1

4.88

Step-by-step explanation:

Formic acid is a weak acid and its dissociation can be represented by the equation: HCOOH ⇌ H+ + HCOO-. The acid dissociation constant (Ka) for this reaction is 1.77*10^-4. When NaOH is added to the solution, it will react with the formic acid to produce its conjugate base (formate ion) and water. This reaction can be represented by the equation: HCOOH + OH- → HCOO- + H2O.

To calculate the resulting pH, we need to determine the number of moles of formic acid and NaOH present in the solution. The number of moles of formic acid can be calculated by multiplying its molarity by its volume in liters: 0.25 M * 0.055 L = 0.01375 moles. Similarly, the number of moles of NaOH can be calculated as: 0.12 M * 0.075 L = 0.009 moles.

Since NaOH is a strong base, it will react completely with formic acid to produce formate ions and water. This means that 0.009 moles of formic acid will react with 0.009 moles of NaOH to produce 0.009 moles of formate ions and water.

After the reaction, there will be 0.01375 - 0.009 = 0.00475 moles of formic acid left in the solution and 0.009 moles of formate ion produced.

The total volume of the solution after adding NaOH is 55 mL + 75 mL = 130 mL or 0.13 L.

We can now use an ICE table to calculate the equilibrium concentrations of all species present in the solution:

(ICE table Picture attached below)

Substituting these values into the expression for Ka and solving for x gives us:

Ka = [H+][HCOO-]/[HCOOH] = x(0.009/0.13 + x)/(0.00475/0.13 - x) = 1.77*10^-4

Solving this quadratic equation gives us x = [H+] = 1.33*10^-5 M.

The pH of the solution can now be calculated as pH = -log[H+] = -log(1.33*10^-5) = 4.88.

So, the resulting pH after adding NaOH to the formic acid solution is 4.88.

Answer 2

To solve this problem, we first need to write the balanced chemical equation for the reaction between formic acid (HCOOH) and sodium hydroxide (NaOH): HCOOH + NaOH → NaCOOH + H2O From the equation, we can see that 1 mole of HCOOH reacts with 1 mole of NaOH to produce 1 mole of NaCOOH and 1 mole of H2O. Therefore, the number of moles of NaOH added can be calculated as follows: moles of NaOH = concentration × volume = 0.12 M × 0.075 L = 0.009 mol Since the reaction is a neutralization reaction, the number of moles of HCOOH initially present can be calculated as follows: moles of HCOOH = concentration × volume = 0.25 M ×