Question

Determine the ammonia concentration of an aqueous solution that has a ph of 11.00. the equation for the dissociation of nh3 (kb = 1.8 × 10-5) is determine the ammonia concentration of an aqueous solution that has a ph of 11.00. the equation for the dissociation of nh3 (kb = 1.8 × 10-5) is 1.8 × 10-2 m

Answer 1

Answer: The concentration of ammonia is 0.056 M

Explanation:

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 11.00

So, $pOH=14-11.00=3.00$

To calculate hydroxide ion concentration, we use the equation:

$pOH=-\log[OH^-]$

We know that:

pOH = 3.00

$3.00=-\log [OH^-]$

$[OH^-]=10^{-3}M$

The chemical equation for the hydrolysis of ammonia follows:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

Let the concentration of ammonia be 'x'

The expression of $K_b$ for above equation follows:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

We are given:

$K_b=1.8\times 10^{-5}$

$[NH_4^+]=[OH^-]=10^{-3}M$

Putting values in above expression, we get:

$1.8\times 10^{-5}=\frac{10^{-3}\times 10^{-3}}{x}\\\\x=\frac{10^{-6}}{1.8\times 10^{-5}}=0.056M$

Hence, the concentration of ammonia is 0.056 M

Answer 2

Taking into account the definition of pH, pOH and kb, the concentration of ammonia is 0.056 M.

First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

The following relationship can be established between pH and pOH:

pOH + pH= 14

Being pH= 11, pOH is calculated as:

pOH + 11= 14

pOH= 14 - 11

pOH= 3

Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:

- log [OH⁻]= 3

Solving:

[OH⁻]= 10⁻³  M

The chemical equation for the hydrolysis of ammonia follows:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Let the concentration of ammonia be 'x'.

On the other side, the dissociation constant of a base, Kb, (or basicity constant, or basic ionization constant) is a measure of the strength of a weak base. So the base dissociation constant is a measure of how completely a base dissociates into its component ions in water.

The dissociation constant Kb is written as the quotient of equilibrium concentrations. In this case:

$Kb=\frac{[NH_{4}^{+} ][OH^{-} ]}{[NH_{3} ]}$

Then, you know:

• Kb=1.8×10⁻⁵
• [NH₄⁺]= [OH⁻]= 10⁻³ M
• [NH₃]= x

Replacing:

$1.8x10^{-5} =\frac{10^{-3}10^{-3} }{x}$

Solving:

$1.8x10^{-5} =\frac{10^{-6} }{x}$

$x =\frac{10^{-6} }{1.8x10^{-5}}$

x= 0.056 M

Finally, the concentration of ammonia is 0.056 M.

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