What is the $p H$ of a saturated solution of $X (O H)_{3}$ will be if its solubility product = $2.7 \times 10^{- 43}$ ?
[Given: log 3 = 0.5, log 2 = 0.3]

10.5

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

7

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

3.5

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

3.8

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

Open in App

Solution

The correct option is D $3.5$
$X {(O H)}_{3} ⟶ X^{+} + 3 {O H}^{-}$
$K_{s p} = [X^{+}] {[{O H}^{-}]}^{3}$
$2.7 \times 10^{- 43} = s {(3 s)}^{3} = {27 s}^{4}$
$s = \sqrt[4]{K_{s p} / 27} = \sqrt[4]{\frac{2.7 \times 10^{- 43}}{27}}$
$= \sqrt[4]{10^{- 44}} = {(10^{- 44})}^{1 / 4} = 10^{- 11}$
$∴$ $[{O H}^{-}] = 3 \times 10^{- 11} M = p O H = - l o g (3 \times 10^{- 11})$
$p O H = - (l o g 3 + l o g 10^{- 11}) = 10.5$
$∴$ $p H = 14 - 10.5 = 3.5$

Suggest Corrections

Similar questions

Q. The solubility product of

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

5 \times 10^{- 19}

25^{o} C

, pH of saturated solution of

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

will be:

Q. The

p H

of a solution containing

0.4 M

H C O_{3}^{-}

and

0.2 M

C O_{3}^{2 -}

is:

[K_{a_{1}} (H_{2} C O_{3}) = 4 \times 10^{- 7}; K_{a_{2}} (H C O_{3}^{-}) = 4 \times 10^{- 11}]

(G i v e n, l o g 4 = 0.60, l o g 0.5 = - 0.3)

A current of 0.250 A is passed through 400 ml of a 2.0 M solution of NaCl for 32 minutes and 10 seconds. What will be the pH of the solution after the current is turned off?
Take log 2 = 0.3, log 3 = 0.5, log 5 = 0.7

pH of a saturated solution of $B a (O H))_{2}$ is $12$ . Its solubility product is -

N a O H

is a strong base. What will be the

p H

5.0 \times 10^{- 2} M N a O H

solution?

(log 2 = 0.3)

Join BYJU'S Learning Program

What is the pH of a saturated solution of X(OH)3 will be if its solubility product = 2.7×10−43?[Given: log 3 = 0.5, log 2 = 0.3]

The correct option is D 3.5X(OH)3⟶X++3OH−Ksp=[X+][OH−]32.7×10−43=s(3s)3=27s4s=4√Ksp/27=4√2.7×10−4327 =4√10−44=(10−44)1/4=10−11∴ [OH−]=3×10−11M=pOH=−log(3×10−11)pOH=−(log3+log10−11)=10.5∴ pH=14−10.5=3.5

What is the $p H$ of a saturated solution of $X (O H)_{3}$ will be if its solubility product = $2.7 \times 10^{- 43}$ ?
[Given: log 3 = 0.5, log 2 = 0.3]