1. Tardigrade
  2. Question
  3. Chemistry
  4. Given that (in ohm -1 cm 2 eq -1 ); T =298 K λ E ∞ for Ba ( OH )2=228.8 specific conductance λ E ∞ for BaCl 2=120.3 for 0.2 N NH 4 OH solution λ E ∞ for NH 4 Cl =129.8 is 4.766 × 10-4 ohm -1 cm -1 then value of pH of the given solution of NH 4 OH will be nearly

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Q. Given that (in ohm $^{-1} cm ^{2} eq ^{-1}$ ); $T =298\, K$
$\lambda_{ E }^{\infty}$ for $Ba ( OH )_{2}=228.8$ specific conductance
$\lambda_{ E }^{\infty}$ for $BaCl _{2}=120.3$ for $0.2\, N\, NH _{4} OH$ solution
$\lambda_{ E }^{\infty}$ for $NH _{4} Cl =129.8$ is $4.766 \times 10^{-4} ohm ^{-1} cm ^{-1}$
then value of $pH$ of the given solution of $NH _{4} OH$ will be nearly

Electrochemistry

Solution:

$\lambda_{ E }^{\infty} \text { for } NH _{4} OH =228.8-120.3+129.8=238.3$
Now, $\lambda_{ E }=\frac{1000 \times k}{N}$
$=\frac{1000 \times 4.766 \times 10^{-4}}{0.2}=2.383$
$\therefore \alpha=\frac{\lambda_{E}}{\lambda_{E}^{\infty}}=\frac{2.383}{238.3}=0.01$
$\therefore OH ^{-}=\alpha \times C =.01 \times 0.2=2 \times 10^{-3}$
$\therefore pOH =3-\log 2=3-0.3=2.7$
$\therefore pH =14-2.7=11.3$