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

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Step-by-step explanation:

Formic acid (HCOOH) reacts with sodium hydroxide (NaOH) to form sodium formate (HCOONa) and water. The balanced chemical equation is:

HCOOH + NaOH → HCOONa + H2O

The reaction is a strong acid-strong base titration. We can use the following equation to calculate the concentration of formate ion (HCOO^-) in the resulting solution:

[HCOO^-] = [OH^-] - [HCOOH]

where [OH^-] is the concentration of hydroxide ion and [HCOOH] is the concentration of formic acid before the reaction.

Before the reaction, the solution contains 0.25 mol/L of formic acid in 55.0 mL, or 0.25 mol/L × 0.055 L = 0.01375 mol of formic acid. The solution also contains 0.12 mol/L of sodium hydroxide in 75.0 mL, or 0.12 mol/L × 0.075 L = 0.009 mol of sodium hydroxide.

Since the reaction between formic acid and sodium hydroxide is a 1:1 reaction, all the 0.009 mol of sodium hydroxide will react with 0.009 mol of formic acid, leaving 0.00475 mol of formic acid unreacted.

[HCOO^-] = [OH^-] - [HCOOH]

[OH^-] = [NaOH] = 0.12 mol/L × 0.075 L / 0.13 L = 0.0692 mol/L

[HCOO^-] = 0.0692 mol/L - 0.00475 mol/L = 0.0645 mol/L

Now we can calculate the pH of the resulting solution using the Ka expression for formic acid:

Ka = [HCOO^-][H3O^+]/[HCOOH]

[H3O^+] = Ka × [HCOOH] / [HCOO^-]

[H3O^+] = 1.77 × 10^-4 × 0.00475 mol/L / 0.0645 mol/L

[H3O^+] = 1.29 × 10^-5 mol/L

pH = -log[H3O^+]

pH = -log(1.29 × 10^-5)

pH = 4.89

Therefore, the resulting pH is 4.89.