The equilibrium constant for the ionization of RN H 2 g in water as RNH 2 g - H 2 O l - RNH 3- aq - OH - aq is 8 - 106-6 at 25- C-Find the pH of a solution at equilibrium when pressure of RN H 2 g is 0-5 bar log 2-0-3- log 3-0-7A- 12-3B- 11-3C- 11-47D- 6-7

The correct option is B 11.3K = (x)(x)/0.5 8×10^ 6 = (x)^2/0.5 x^2 = 4×10^ 6 x nbsp;= 2×10^ 3 pOH = log(2×10^ 3 nbsp; nbsp; nbsp; nbsp; nbsp; ...

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

Byju's Answer

Standard IX

Physics

Thrust and Pressure

The equilibri...

Question

The equilibrium constant for the ionization of $R N H_{2} (g)$ in water as
$R N H_{2} (g) + H_{2} O (l) ⇌ R N H_{3}^{+} (a q) + O H^{-} (a q)$
is $8 \times 106^{- 6}$ at $25^{\circ} C$ .Find the pH of a solution at equilibrium when pressure of $R N H_{2} (g)$ is 0.5 bar
$(l o g 2 = 0.3, l o g 3 = 0.7)$ :

12.3

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

11.3

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

11.47

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

6.7

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

Open in App

Solution

The correct option is B 11.3
$K = \frac{(x) (x)}{0.5}$
$8 \times 10^{- 6} = \frac{(x)^{2}}{0.5}$
$x^{2}$ = $4 \times 10^{- 6}$
$x = 2 \times 10^{- 3}$

$p O H = - l o g (2 \times 10^{- 3}$
$= 2.7$
$p H = 14 - 2.7 = 11.3$

Suggest Corrections

Similar questions

Q. The equilibrium constant for the ionization of

R {N H}_{2} (g)

in water as

R {N H}_{2} (g) + H_{2} O (l) ⇌ R {N H}_{3}^{+} (a q) + {O H}^{-} (a q)

{8 \times 10}^{- 6}

25^{o} C

. Find the pH of a solution at equilibrium when pressure of

R {N H}_{2} (g)

0.5

bar:

Q. The equilibrium constant for the ionization of

R N H_{2} (g)

in water :

R N H_{2} (g) + H_{2} O (l) ⇌ R N H_{3}^{+} (a q) + O H (a q)

10^{- 6}

25^{o} C

. Select the correct statement(s) :

Q. A hand book states that the solubility of

R N H_{2} (g)

in water at

1

atm and

25^{\circ} C

22.41

volumes of

R N H_{2} (g)

per volume.

(p K_{b} o f R N H_{2} = 4)

. Find the maximum

p O H

that can be attained by dissolving

R N H_{2}

in water.

Q. When 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M

C H_{3} C O O H

solution, the pH will be:
(use :

(p K_{a})_{a c e t i c a c i d} = 5, l o g 2 = 0.3, l o g 3 = 0.5, l o g 5 = 0.7)

Q. A hand book states that the solubility of

R N H_{2} (g)

in water at

1

atm and

0^{o}

C is

22.41

volumes of

R N H_{2} (g)

per volume of water. (

p K_{b}

R N H_{2} = 4

) Find the max. pOH that can be attained by dissolving

R N H_{2}

in water.

Join BYJU'S Learning Program

The equilibrium constant for the ionization of RNH2(g) in water as RNH2(g)+H2O(l)⇌RNH+3(aq)+OH−(aq) is 8×106−6 at 25∘C.Find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar (log 2=0.3,log 3=0.7) :

The correct option is B 11.3K = (x)(x)0.5 8×10−6 = (x)20.5 x2 = 4×10−6 x=2×10−3 pOH=−log(2×10−3 =2.7 pH=14−2.7=11.3

The correct option is B 11.3
$K = \frac{(x) (x)}{0.5}$
$8 \times 10^{- 6} = \frac{(x)^{2}}{0.5}$
$x^{2}$ = $4 \times 10^{- 6}$
$x = 2 \times 10^{- 3}$

$p O H = - l o g (2 \times 10^{- 3}$
$= 2.7$
$p H = 14 - 2.7 = 11.3$