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6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
A buffer solution is prepared in which the concentration of $NH_3$ is $0.30\, M$ and the concentration of $NH_4^+$ is $0.20\, M.$ If the equilibrium constant, $K_b$ for $NH_3$ equals $1.8 \times 10^{-5},$ what is the $pH$ of this solution ?  $(log \, 2.7 = 0.43)$
  • A
    $9.08$
  • ✓
    $9.43$
  • C
    $11.72$
  • D
    $8.73$
✓

Answer

Correct option: B.
$9.43$
b
$\mathrm{pOH}=\mathrm{pK} \mathrm{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$

$=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$

$=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}$

$=-5-0.25+(-0.176)$

therefore,

$4.75-0.176=4.57$

therefore, $\mathrm{pH}=14-4.57=9.43$

$\mathrm{pOH}=\mathrm{pK} \mathrm{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$

$=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$

$=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}$

$=-5-0.25+(-0.176)$

therefore,

$4.75-0.176=4.57$

therefore, $\mathrm{pH}=14-4.57=9.43$

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