Acid-Base Equilibrium Questions 1. Calculate the equilibrium concentrations of H

Question

Acid-Base Equilibrium Questions 1. Calculate the equilibrium concentrations of HF. HO and F at equalibcium if the initial concentration of HF is 0.0250 MCalculate the pH and poH. (K for HF is 7-2 x 10") 2. Calculate the equilbrium concentrations of HNO. HO and NO; at equilibrium if the initial concentration of HNO2 is 1.00 M. Calculate the pH and poll. (K for HNC) is 5.1 10 3. Calculate the equilbsium concentrations of phenol, C.H.OH, HO" and CHO at equilibrium if the initial concentration of C.H OH is 0.200 M. Calculate the phi and pOH. (K, for C.H OH is 1.0x 100) Calculate the equilibrium concentrations of NHs, NH. and OH at equilibrium if the initial concentration of NH, ?s 1.00 M. Calculate the pH and pOH. (K for NII, is 1.8 x 10) 4. Calculate the equilibrium concentrations of pyridine, C,H.N, the pyridinium ion, G,HNH', and OH at equilibrium if the initial concentration of CsHsN is 0.100 M. Calculate the pH and pOH. (K for CsHsN is 1.7 x 10) 5. 6. Calculate the equilibrium concentrations of ethanolamine (HOCH CH:NH), ethanolammonium ion, HOCH CH:NHs, and OH at equilibrium if the initial concentration of HOCH CH-NH, is 0.0200 M. Calculate the pH and pOH. (K, for HOCHCH2NH, ?s 3.3 x 10") CGalor

Explanation / Answer

ANSWER:

Us ICE procedure to the equilibrium to solve the problem

HOCH2CH2NH2 + H2O <--------> HOCH2CH2NH3+ + OH-

I 0.0200M 0 0

C -xM +x +x

E     (0.0200 - x)M +x +x

Kb = [HOCH2CH2NH3+ ] [ OH-] / [HOCH2CH2NH2]

The concentration of water [H2O] is taken as unity, because it acts as solvent and is present in excess. Its concentration does not change practically.

3.3 X 105 = (x) X (x) / (0.0200 - x)

3.3 X 10-5 X (0.0200 - x) = x2

6.6 X 10-7 - 3.3 X 10-5 x = x2

x2 + 3.3 X 10-5 x - 6.6 X 10-7 = 0

Above equation is quadratic equation its solution gives us the value of x,

x = 8.0 X 10-4

Hence [HOCH2CH2NH3+ ] = [ OH-] = 8.0 X 10-4M

[HOCH2CH2NH2] = (0.0200 - x) = (0.0200 - 8.0 X 10-4) = 0.0192M

pOH = -log[ OH-] = - log(8.0 X 10-4M) = 3.09

pH + pOH = 14

pH = 14 - 3.09 = 10.91

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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