The OH -concentration in a mixture of 5.0 mL of 0.0504 M NH 4 Cl and 2 mL of 0.0210 M NH 3 solution is x × 10-6 M. The value of x is . (Nearest integer) [. Given K w =1 × 10-14 and .K b =1.8 × 10-5]

Question

The OH -concentration in a mixture of 5.0 mL of 0.0504 M NH 4 Cl and 2 mL of 0.0210 M NH 3 solution is x × 10-6 M. The value of x is . (Nearest integer) [. Given K w =1 × 10-14 and .K b =1.8 × 10-5] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

[ NH 4+]=0.0504 [ NH 3]=0.0210 So K b =([ NH 4+][ HO -]/[ NH 3]) [ HO -]=( K b ×[ NH 3]/[ NH 4+]) =1.8 × 10-5 × (2/5) × (210/504) =3 × 10-6

Q. The $O H^{-}$ concentration in a mixture of $5.0 m L$ of $0.0504 M N H_{4} Cl$ and $2 m L$ of $0.0210 M N H_{3}$ solution is $x \times 1 0^{- 6} M$ . The value of $x$ is ________. (Nearest integer)
$[$ Given $K_{w} = 1 \times 1 0^{- 14}$ and $K_{b} = 1.8 \times 1 0^{- 5}]$

$[N H_{4}^{+}] = 0.0504 & [N H_{3}] = 0.0210$
So $K_{b} = \frac{[ N H _{4}^{+} ] [ H O ^{-} ]}{[ N H _{3} ]}$
$[H O^{-}] = \frac{K _{b} \times [ N H _{3} ]}{[ N H _{4}^{+} ]}$
$= 1.8 \times 1 0^{- 5} \times \frac{2}{5} \times \frac{210}{504}$
$= 3 \times 1 0^{- 6}$

Q. The OH−concentration in a mixture of 5.0mL of 0.0504MNH4​Cl and 2mL of 0.0210MNH3​ solution is x×10−6M. The value of x is ________. (Nearest integer) [ Given Kw​=1×10−14 and Kb​=1.8×10−5]

[NH4+​]=0.0504&[NH3​]=0.0210 So Kb​=[NH3​][NH4+​][HO−]​ [HO−]=[NH4+​]Kb​×[NH3​]​ =1.8×10−5×52​×504210​ =3×10−6

Q. The $O H^{-}$ concentration in a mixture of $5.0 m L$ of $0.0504 M N H_{4} Cl$ and $2 m L$ of $0.0210 M N H_{3}$ solution is $x \times 1 0^{- 6} M$ . The value of $x$ is ________. (Nearest integer)
$[$ Given $K_{w} = 1 \times 1 0^{- 14}$ and $K_{b} = 1.8 \times 1 0^{- 5}]$

$[N H_{4}^{+}] = 0.0504 & [N H_{3}] = 0.0210$
So $K_{b} = \frac{[ N H _{4}^{+} ] [ H O ^{-} ]}{[ N H _{3} ]}$
$[H O^{-}] = \frac{K _{b} \times [ N H _{3} ]}{[ N H _{4}^{+} ]}$
$= 1.8 \times 1 0^{- 5} \times \frac{2}{5} \times \frac{210}{504}$
$= 3 \times 1 0^{- 6}$