Calculate the pH of 0.005M $B a (O H)_{2}$ in aq. solution.

Open in App

Solution

Barium hydroxide is a strong base and completely dissociated.

$B a (O H)_{2} \to B a^{2 +} + 2 O H^{-}$

So, $0.005 M B a (O H)_{2}$ yields $0.005 \times 2 = 0.01$ moles of $O H^{-}$

So, $p O H = - l o g [O H^{-}] = log \frac{1}{10^{- 2}} = log 10^{2} = 2$

Now, as we know

$p H + p O H = 14$

$p H = 14 - 2 = 12$

Suggest Corrections

Similar questions

0.005 M

B a {(O H)}_{2}

25^{o} C

100 m l

1000 m l

The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH)₂ is:

p H

(B a {(O H)}_{2})

p H

50 m L

0.01 M

B a {(O H)}_{2}

50 m L

Join BYJU'S Learning Program