Question

pH and pOH calculations for weak acids and weak bases

1. What is the pH of a 0.55 M solution of trimethylamine (pKb= 4.13)

2. What is the pH of a 0.25 M solution of jaspersamine? The pKb value for jaspersamine 4.40.

3. What is the pH of a 0.15 M solution of a weak acid ammonium bromide given that the Kb value for ammonia is 1.8x10^-5.

Answer 1

Following is the - complete Answer -&- Explanation: for Question - 1 and Question - 2, of the given: Question Set:... in....typed format...

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$\Rightarrow$Question - 1:

$\Rightarrow$Answer:

For tri-methyl-amine:   pH = 11.805

$\Rightarrow$Explanation:

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

• Given:

1. Molarity of Tri-methyl amine: M_a = 0.55 M ( mol/L )
2. For  Tri methyl amine: value of pK_b = 4.13

• ​​​​​​​Step - 1:

​​​​​​​$\Rightarrow$  pK_b = 4.13

$\Rightarrow$K_b = 10 ^-4.13 = 7.413 x 10^-5   

• Step - 2:

​​​​​​​$\Rightarrow$We know: K_b = [X⁺ ] x [OH ^- ] / [ tri-methyl amine ] = 7.413 x 10^-5   

$\Rightarrow$ [OH ^- ] = $\sqrt{[Kb * [trimethyl amine]]}$ = (
[(7.413 10 5)ax(0.55)]
) =  0.00638 M ( mol/L )

• Step - 3:

​​​​​​​$\Rightarrow$pOH = - log ₁₀ [OH^- ] = - log₁₀ [ 0.00638 ] = 2.195

$\Rightarrow$ pH =  ( 14.0 - pOH ) = ( 14.0 - 2.195 ) = 11.805

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$\Rightarrow$Question - 2:

$\Rightarrow$Answer:

For jaspersamine:   pH = 11.5

$\Rightarrow$Explanation:

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

• Given:

1. ​​​​​​​Molarity of jaspersamine: M_a = 0.25 M ( mol/L )
2. For  jaspersamine: value of pK_b = 4.40

• ​​​​​​​Step - 1:

​​​​​​​$\Rightarrow$  pK_b = 4.40

$\Rightarrow$K_b = 10 ^-4.40 = 3.98 x 10 ^-5   

• Step - 2:

​​​​​​​$\Rightarrow$We know: K_b = [X⁺ ] x [OH ^- ] / [ jaspersamine ] = 3.98 x 10 ^-5

$\Rightarrow$ [OH ^- ] = $\sqrt{[Kb * [jaspersamine]]}$ = ( $\sqrt{[(3.98 * 10^{-5} ) * ( 0.25 ) ]}$ ) =  0.00315 M ( mol/L )

• Step - 3:

​​​​​​​$\Rightarrow$pOH = - log ₁₀ [OH^- ] = - log₁₀ [ 0.00315 ] = 2.50

$\Rightarrow$ pH =  ( 14.0 - pOH ) = ( 14.0 - 2.50 ) = 11.5  

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Note: The remaining question: i.e. Question - 3: can be solved, by applying the method followed for solving the above stated solutions...

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