Home
Class 11
CHEMISTRY
A solution consist of 0.2 M NH(4)OH and ...

A solution consist of `0.2 M NH_(4)OH` and `0.2 M NH_(4)Cl`. If `K_(b)` of `NH_(4)OH` is `1.8 xx 10^(-5)`, the `[OH^(-)]` of the resulting is

A

`0.9xx10^(-5)M`

B

`1.8xx10^(-5)M`

C

`3.2xx10^(-5)M`

D

`3.6xx10^(-5)M`

Text Solution

Verified by Experts

The correct Answer is:
B
`[OH^(-)]=(K_(b)["Base"])/(["Salt"])`
`=1.8xx10^(-5)xx(0.2)/(0.2)=1.8xx10^(-5)`
Promotional Banner

Topper's Solved these Questions

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (Solubility Product)|5 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (H.W.) (IONISATION OF ACIDS & BASES, DEGREE OF IONISATION & IONIC PRODUCT)|6 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-II (Salt Hydrolysis)|3 Videos

  • 14TH GROUP ELEMENTS

    NARAYNA|Exercise Subjective Questions|6 Videos

  • ALKANES

    NARAYNA|Exercise Integer Answer Type|7 Videos

Similar Questions

Explore conceptually related problems

A 1L solution contains 0.2M NH_(4)OH and 0.2M NH_(4)Cl. If 1.0 mL of 0.001 M HCl is added to it what will be the [OH^(-)] of the resulting solution (K_(b)=2xx10^(-5))

Solution of 0.1 N NH_(4)OH and 0.1 N NH_(4)Cl has pH 9.25 , then find out K_(b) of NH_(4)OH .

K_(b) for NH_(4)OH is 1.8 xx 10^(-5) . The [overset(Theta)OH] of 0.1 M NH_(4)OH is

Calculate pH of the buffer solution containing 0.15 mole of NH_(4)OH " and " 0.25 mole of NH_(4)Cl. K_(b) " for " NH_(4)OH " is " 1.98 xx 10^(-5) .

100ml of 0.75 N NH_(4)OH is mixed with 100ml of 0.25 N HCl. The K_(b) of NH_(4) OH = 2 xx 10^(-5) . Hence pH of solution is .