Question

4. Calculate [H30*], [OH-], pH and pOH of a 0.5 M CH3COOH solution, K. = 1.8 x 10-5. 5. Write the acid ionization equations in water of the weak acid H PO4, and the expressions of K.

Answer 1

Following is the - complete Answer -&- Explanation: for the first question ( i.e. Question - 4 ), of the given: Question Set......in...typed format...

$\Rightarrow$Answer:

As we have calculated: following are the answers:

1. [ H₃O⁺] = 0.002991 M ( mol/L )
2. [ OH^- ] =   3.34 x 10^-12 M ( mol/L )
3. pH =   2.524
4. pOH = 11.476 ​​​​​​​

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answers.

• Given:

1. ​​​​​​​Initial molar concentration of acetic acid: [CH₃COOH] = 0.5 M ( mol/L )
2. For acetic acid: acid dissociation constant: K_a = 1.8 x 10^-5

• ​​​​​​​Step - 1:

​​​​​​​Following is the balanced chemical equation: for the dissociation of the aqueous solution of acetic acid: into ions...

$\Rightarrow$ CH₃COOH (aq) + H₂O (l) $\rightleftharpoons$ H₃O⁺ (aq) + CH₃COO^- (aq) ------------------------Equation - 1

$\Rightarrow$We know: initial concentration: of acetic acid:  [CH₃COOH] = 0.5 M ( mol/L )

• Step - 2:

​​​​​​​Therefore, based on Equation - 1, and the initial concentration of acetic acid: we can form the following ICE Table...

	[CH3COOH], M	[ H3O+], M	[CH3COO-], M
Initial ( concentration )	0.5	0.0	0.0
Change ( concentration)	- X	+ X	+ X
Equilibrium ( concentration )	(0.5 - X )	+ X	+ X

Therefore: we will find the following concentrations, at equilibrium:

1. [CH₃COOH]_eq =  ( 0.5 - X ) M
2. [ H₃O⁺]_eq =  + X (mol/L )
3. [CH₃COO^-]_eq = + X (mol/L )

• ​​​​​​​Step - 3:

​​​​​​​Therefore: we will get the following:

$\Rightarrow$K_a = [ H₃O⁺]_eq x  [CH₃COO^-]_eq / [CH₃COOH]_eq =   1.8 x 10^-5

OR,

$\Rightarrow$   K_a = ( X ) x ( X ) / [ ( 0.5 - X ) ] =   1.8 x 10^-5

$\Rightarrow$ K_a = ( X² ) / [ ( 0.5 - X ) ] =   1.8 x 10^-5

$\Rightarrow$ X² + ( 1.8 x 10^-5 ) X - ( 9.0 x 10^-6) = 0.0 ----------------------------Equation - 2

$\Rightarrow$Therefore: solving Equation - 2, we would get the following solution:

i.e.

$\Rightarrow$ X = 0.002991 M ( mol/L )

• Step - 4:

​​​​​​​Therefore: we will find the following:

1. [ H₃O⁺]_eq =  + X (mol/L ) = 0.002991 M ( mol/L )
2. [CH₃COO^-]_eq = + X (mol/L ) = 0.002991 M ( mol/L )
3. [CH₃COOH]_eq =  ( 0.5 - X ) M = 0.497 M ( mol/L )

• ​​​​​​​Step - 5:

​​​​​​​Therefore: since we know:

$\Rightarrow$  pH = - log [H₃O⁺] = - log [ 0.002991 ] =  2.524

Since we know:  pH + pOH = 14, we will find the following:

$\Rightarrow$pOH =  ( 14.0 - pH ) = 14.0 - 2.524 = 11.476

$\Rightarrow$ Since: pOH = - log [ OH^- ]

$\Rightarrow$   [ OH^- ] = 10^-pOH = 10 ^-11.476 = 3.34 x 10^-12 M ( mol/L )

Therefore: we will get the following Answers:

• Answers:

1. ​​​​​​​[ H₃O⁺] = 0.002991 M ( mol/L )
2. [ OH^- ] =   3.34 x 10^-12 M ( mol/L )
3. pH =   2.524
4. pOH = 11.476 ​​​​​​​