What is the conjugate acid of NH_3 ? NH_3 NH_2^+ NH_3^+ NH_4^+ NH_4 OH for a buf

Question

What is the conjugate acid of NH_3 ? NH_3 NH_2^+ NH_3^+ NH_4^+ NH_4 OH for a buffer prepared by combining 0 370 mole of formic acid (HCOOH) and 0.230 mole of sodium format (NaCOOH) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 times 10^-4, use the above information to answer the following questions. Write the two reactions taking place in the buffer, check all reactions are balanced. Identify the common ion in the above two reactions. Write the equilibrium expression for the weak acid. State the approximation you would use to calculate the pH of the buffer?

Explanation / Answer

Dear student,

a) Two reactions are given below,

HCOONa(aq)<==> Na+(aq) + HCOO-(aq)

HCOOH(aq) <==> H+(aq) + HCOO-(aq)

HCOOH(aq) + H2O(l)<==> H3O+(aq) + HCOO-(aq)...........................................(1)

b)

HCOONa(aq)<==> Na+(aq) + HCOO-(aq)

HCOOH(aq) <==> H+(aq) + HCOO-(aq)

Therefore the common ion in above two reaction is HCOO-(aq).

c) Equilibrium constant expression for weak acid,(formic acid is a weak acid)

Ka= [H3O+]*[HCOO-] / [HCOOH] = 1.77*10–4 (given) .....................(2)

d) Calculation of pH of the buffer solution

HCOOH(aq)    H+(aq) or  H3O+(aq) HCOO-(aq)

Initial 0.370 x 0.230

Change -x +x +x

Equilibrium 0.370-x x 0.230+x   

Therefore from equation 2

Ka = [H3O+]*(0.230+x) / (0.370-x)   [H3O+]*(0.230) / (0.370)

or,   [H3O+] = 1.77*10–4 *(0.370) / (0.230)=2.847*10–4 M

Therefore

pH = –log[H3O+]

pH= –log[2.847*10–4]

pH=3.54 M

N.B : You are also calculate pH by  H+(aq) instead of  [H3O+]

Regards,

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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