Question

Question : A 0.010Msolution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010Msolution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.

(a) Which acid has the larger equilibrium constant for ionization

HA[HA(aq) ⇌ A⁻(aq) + H⁺(aq)]or HB[HB(aq) ⇌ H⁺(aq) + B⁻(aq)]?

(b) What are the equilibrium constants for the ionization of these acids?

(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A⁻ or B⁻), and the hydrogen ion (H⁺). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Short answer

The Result is

1. HB Solution has larger equilibrium constant
2. The Equillibrium constant for HA is \(5 \times {10^{ - 4}}\)

The Equillibrium Constant for HB is \(2.85 \times {10^{ - 3}}\)

Step-by-step solution

Definition

Ionization or ionization is the process by which anatom or amolecule acquires a egative or positivecharge by gaining or losingelectrons, often in conjunction with other chemical changes.

Weak acids concentration and osmotic pressure

Given information

The Concentration of HA is\(0.010{\rm{M}}\)

The Osmotic pressure of HA solution is\(0.293{\rm{atm}}\)

The Concentration of HB is\(0.010{\rm{M}}\)

The Osmotic pressure of HB solution is\(0.345{\rm{atm}}\)

The Temperature is\({25^ \circ }{\rm{C}}(25{\rm{K}} + 273{\rm{K}} = 298{\rm{K}})\)

The ideal gas constant \({\rm{R}} = 0.08206{\rm{Latm}}/{\rm{mol}}\)

To find larger equilibrium constant for ionization

In order to find that, We have to use the osmotic pressure formula

\({\rm{\Pi }} = iMRT\)

Since the van’ t Hoff factor (i) equals

\(i = \frac{{{\rm{\;moles of particles in solution\;}}}}{{{\rm{\;moles of formula units dissolved\;}}}}\)

The solution with greater van’t Hoff factor, will have the larger equilibrium constant (because\(K = \frac{{{\rm{\;Products\;}}}}{{{\rm{\;Reactants\;}}}}\), and the larger value of i, larger the product)

-i value of HA Solution

\(\begin{array}{*{20}{c}}{{\rm{\Pi }} = iMRT}\\{i = \frac{{\rm{\Pi }}}{{MRT}}}\\{ = \frac{{0.293}}{{0.010 \times 0.08206 \times 298}}}\\{ = 1.20}\end{array}\)

-i value for HB Solution

\(\begin{array}{*{20}{c}}{{\rm{\Pi }} = iMRT}\\{i = \frac{{\rm{\Pi }}}{{MRT}}}\\{ = \frac{{0.345}}{{0.010 \times 0.08206 \times 298}}}\\{ = 1.41}\end{array}\)

Since HB Solution has the greater value of vant Hoff factor, it has also the larger equilibrium constant

The Equillibrium constants for the ionization of HA

HA

\({\rm{HA}}({\rm{aq}}) \to {{\rm{A}}^ - }({\rm{aq}}) + {{\rm{H}}^ + }({\rm{aq}})\)

Initial (M)	\(0.01\)	\(0\)	\(0\)
Change(M)	\( - {\rm{x}}\)	\( + x\)	\( + x\)
Equillibrium(M)	\(0.01 - x\)	\(X\)	\(X\)

The moles of particles in solution\( = 0.01 - x + x + x\)

The moles of formula units dissolved\( = 0.01\)

Therefore, the Value of\(x\)is

\(\begin{array}{*{20}{c}}{i = \frac{{{\rm{\;moles of particles in solution\;}}}}{{{\rm{\;moles of formula units dissolved\;}}}}}\\{1.20 = \frac{{0.01 - x + x + x}}{{0.01}}}\\{x = 0.002}\end{array}\)

Now we will calculate the equilibrium constant of HA

\(\begin{array}{*{20}{c}}{{K_{HA}} = \frac{{\left( {{A^ - }} \right) \times \left( {{H^ + }} \right)}}{{(HA)}}}\\{ = \frac{{x \times x}}{{0.01 - x}}}\\{ = \frac{{0.002 \times 0.002}}{{0.01 - 0.002}}}\\{ = 5 \times {{10}^{ - 4}}}\end{array}\)

The Equillibrium constants for the ionization of HB

HB

\({\rm{HB}}({\rm{aq}}) \to {{\rm{B}}^ - }({\rm{aq}}) + {{\rm{H}}^ + }({\rm{aq}})\)

Initial(M)	\(0.01\)	\(0\)	\(0\)
Change(M)	\( - X\)	\( + x\)	\( + x\)
Equillibrium(M)	\(0.01 - x\)	\({\rm{X}}\)	\({\rm{X}}\)

The moles of particles in solution\( = 0.01 - x + x + x\)

The moles of formula units dissolved\( = 0.01\)

Therefore, the Value of\(x\)is

\(\begin{array}{*{20}{c}}{i = \frac{{{\rm{\;moles of particles in solution\;}}}}{{{\rm{\;moles of formula units dissolved\;}}}}}\\{1.41 = \frac{{0.01 - x + x + x}}{{0.01}}}\\{x = 0.0041}\end{array}\)

Now, we will calculate the equilibrium constant of HB

\(\begin{array}{*{20}{c}}{{K_{HB}} = \frac{{\left( {{B^ - }} \right) \times \left( {{H^ + }} \right)}}{{(HB)}}}\\{ = \frac{{x \times x}}{{0.01 - x}}}\\{ = \frac{{0.0041 \times 0.0041}}{{0.01 - 0.0041}}}\\{ = 2.85 \times {{10}^{ - 3}}}\end{array}\)

Therefore,

a) HB Solution has larger equilibrium constant

b) The Equillibrium constant for HA is \(5 \times {10^{ - 4}}\)

The Equillibrium Constant for HB is \(2.85 \times {10^{ - 3}}\)