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

When the dissociation equation of NH3 is:

NH3(aq) + H2O ↔ NH4(aq)  +   OH-(aq)


when  Kb  =(1.8 x 10^-5)

and when PH+POH= 14 

∴ POH = 14 -11= 3

when POH = -㏒[OH]

∴[OH] = 0.001

when [OH] = [NH4+] = 0.001 

when Kb = [NH4+][OH-]/[NH4OH]

by substitution:

1.8 x 10^-5 = (0.001)^2 / [ NH4OH]

∴[NH4OH] = 0.06 M

Answer 2

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