You visited us 1 times! Enjoying our articles? Unlock Full Access!

pH of a Solution

What is the m...

Question

What is the minimum $p H$ when $F e (O H)_{3}$ starts precipitating from a solution contain $0.1 M F e C l_{3}$ ? $K_{s p}$ $F e (O H)_{3} = 8 \times 10^{- 13} M^{3}$

3.7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3.7$
$F e O H_{3} ⟶ F e C l_{3}$
Concentration of $O H^{-}$ after precipitation,
$O H^{-} = \frac{K_{s p} F e (O H)_{3}}{C l^{-}}$
$O H^{-} = \frac{8 \times 10^{- 13}}{.1} = 8 \times 10^{- 12}$
$p O H = 12 - l o g 8 = 11.1$
$p H = 14 - 11.1 = 3.7$

Suggest Corrections

Similar questions

F e {(O H)}_{3}

p H = 8.0

K_{s p}

F e {(O H)}_{3} = 1.0 \times 10^{- 36}

F e {(O H)}_{3}

1 \times 10^{- 36}

{O H}^{-}

F e {(O H)}_{3}

0.001 M

F e {C l}_{3}

0.04 M

F e {C l}_{2}

0.02 M

F e {C l}_{3}

0.01 M

H C l

p H

F e {(O H)}_{2}

F e {(O H)}_{3}

K_{s p}

F e {(O H)}_{2}

F e {(O H)}_{3}

8 \times 10^{- 16}

4 \times 10^{- 38}

0.10 M

M g^{2 +}

M g {(O H)}_{2}

K_{s p}

M g (O H)_{2} = 1.2 \times 10^{- 11} M^{3}

K_{s p}

M g (O H)_{2}

4.0 \times 10^{- 6}

p H,

M g^{2 +}

0.01

M g C I .

Join BYJU'S Learning Program