What is the pH of a solution that is prepared by dissolving 9.96 grams of formic

Question

What is the pH of a solution that is prepared by dissolving 9.96 grams of formic acid (formula weight = 46.03 grams/mol) and 9.07 grams of sodium formate (formula weight = 68.01 grams/mole) in water and diluting to 500.00 mL? The Ka for formic acid is 0.000177. What is the pH of a solution that is prepared by dissolving 9.96 grams of formic acid (formula weight = 46.03 grams/mol) and 9.07 grams of sodium formate (formula weight = 68.01 grams/mole) in water and diluting to 500.00 mL? The Ka for formic acid is 0.000177.

Explanation / Answer

we have

Ka = 0.000177 therefore pKa = - logKa = 3.75

and

moles of formic acid = 9.96 g / 46.03 g/mol = 0.216 mol

moles of sodium formate = 9.07 g / 68.01 g/mol = 0.133 mol

therefore

pH = pKa + log[salt]/[acid]

we have

pH = 3.75 + log(0.1333 / 0.216 )

= 3.75 - 0.21

= 3.54

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.