Question

calculate the concentration of hydroxyl ion in 0.1M solution of ammonium hydroxide having kb = 1.8 × 10^-5 ​

Answer 1

According to the question,

$\sf NH_4OH _{(aq)} \longleftrightarrow NH^+_{4(aq)} + OH^-_{(aq)}$

$\sf K_b = \dfrac{[NH^+_4][OH^-]}{[NH_4OH]} \: \: ...(1)$

Now,

$\sf [NH^+_4]=[OH^-]$

$\sf[NH_4OH] =0.1 M$

substitute NH₄ = OH¯ in equation (1)

$\sf K_b = \dfrac{[OH^-][OH^-]}{[NH_4OH]}$

$\sf K_b = \dfrac{[OH^-]^{2} }{[NH_4OH]}$

$\sf 1.8 \times {10}^{ - 5} = \dfrac{[OH^-]^{2} }{0.1}$

$\sf [OH^-]^{2} = 1.8 \times {10}^{ - 5} \times 0.1$

$\sf [OH^-]^{2} = 1.8 \times {10}^{ - 6}$

$\sf [OH^-] = \big( 1.8 \times {10}^{ - 6} \big)^{ \frac{1}{2} }$

$\sf [OH^-] = 1.34 \times {10}^{ - 3}$

Answer 2

The ammonium hydroxide dissociates as follows:$NH_{4}OH$

$NH_{4 ^{ + } } + OH ^{ - }$

Write the equilibrium expression for the base dissociation:

$k_{b} = [ NH_{4 ^{+} }][OH ^{ - } ]/ NH_{4}OH$

The initial and equilibrium concentrations of the species present:

let x = moles of $NH_{3}$ that dissociate to form $NH_{4 ^{ + } } and {OH}^{ - }$

$NH_{4}OH .... NH_{ {4}^{ + } } .... {Oh}^{ - }$

initial concentrations

$(mol {L}^{ - 1} )$ .... 0.1 ....0 ....0

Equilibrium

concentrations ( mol

0.1-x ... x .... x ....

${L}^{-1})$

Since $NH_{4} OH$ is a weak base($K_{b}$ is small) it dissociates only slightly, x will be very small compared to 0.1

So, at equilibrium, $[NH_{4} OH] = 0.1 mol {L}^{ - 1}$ Substitute the concentration values into the expression for the base dissociation:

$k_{b} = [x][x]/0.1 M$

1.8×${{10}^{-5}}$=${{x}^{2}}$÷0.1 M

x² = 1.8 x${{10}^{-5}}$ x 0.1 M = 1.8 x ${{10}^{-6}}$ M

x = √1.8 x ${{10}^{-6}}$ M = 1.34 x 103 M

[OH] = x= 1.34 x 10-³ M