A buffer solution contains .20 moles of methanoic acid, HCOOH, and .30 moles of

Question

A buffer solution contains .20 moles of methanoic acid, HCOOH, and .30 moles of sodium methanoate, NaCOOH in 1.00 L of the buffer. The acid ionizarion contant, Ka of methanoic acid is 1.8 x 10^-4

a) calculate the pH of the solution

b)Compare the capacity of this buffer to neutalize added acid to its capcity to neutralized added base. Explain your answer completely.

c)If .10 moles of HCL gas solution were bubbled through a liter of the buffer, what would happen to the pH? How would this addition affect the buffer's capcity to neutralized added acid and base in the fute? Answer the question fully, including equations and calcuation where necessary.

Thank You in advance!!

Explanation / Answer

(a) Henderson-Hasselbalch equation:

pH = pKa + log([NaCOOH]/[HCOOH])

= -log Ka + log(moles of NaCOOH/moles of HCOOH)

= -log(1.8 x 10^(-4) + log(0.30/0.20)

= 3.92

(b) Added acid: NaCOOH + H+ => HCOOH + Na+

Acid buffer capacity = moles of NaCOOH = 0.30 mol

Added base: HCOOH + OH- => HCOO- + H2O

Base buffer capacity = moles of HCOOH = 0.20 mol

Thus acid buffer capacity is greater than base buffer capacity

(c) HCl + NaCOOH => HCOOH + NaCl

Acid buffer capacity = moles of NaCOOH = 0.30 - 0.10 = 0.20 mol

Base buffer capacity = moles of HCOOH = 0.20 + 0.10 = 0.30 mol

pH = pH = pKa + log([NaCOOH]/[HCOOH])

= -log Ka + log(moles of NaCOOH/moles of HCOOH)

= -log(1.8 x 10^(-4) + log(0.20/0.30)

= 3.57

Thus the pH will decrease and the base buffer capacity will become greater than acid buffer capacity after adding HCl

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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