Respuesta :
Answer: The concentration of [tex]OH^-[/tex] ions in the solution is 0.0036 M
Explanation:
We are given:
Concentration of ammonia = 0.75 M
The chemical equation for the ionization of ammonia follows:
[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]
Initial: 0.75
At eqllm: 0.75-x x x
The expression of [tex]K_b[/tex] for above equation follows:
[tex]K_b=\frac{[NH^+][OH^-]}{[NH_3]}[/tex]
We know that:
[tex]K_b\text{ for }NH_3=1.74\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]1.74\times 10^{-5}=\frac{x\times x}{(0.75-x)}\\\\x=-0.0036,0.0036[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative
So, concentration of hydroxide ion = 0.0036 M
Hence, the concentration of [tex]OH^-[/tex] ions in the solution is 0.0036 M
The concentration of OH- ions is 0.00367 M.
NH3(aq) + H2O(l) ⇄ NH4^+(aq) + OH^-(aq)
I 0.75 0 0
C -x +x +x
E 0.75 - x x x
The Kb of NH3 is 1.8 * 10^-5
Kb = [ NH4^+] [OH^-]/[NH3]
1.8 * 10^-5 = [x] [x]/[0.75 - x]
1.8 * 10^-5[0.75 - x] = x^2
1.35 * 10^-5 - 1.8 * 10^-5x - x^2 = 0
x^2 + 1.8 * 10^-5x - 1.35 * 10^-5 = 0
x = 0.00367 M
Hence, the concentration of OH- ions is 0.00367 M.
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