Home
Class 11
CHEMISTRY
A solution contains 60 mL of 0.1 M NH(4)...

A solution contains `60 mL` of `0.1 M NH_(4)OH` and `30 mL` of `0.1 MHCl`. The `P^(H)` of the resulting mixture is
(Given `K_(b)` of `NH_(4)OH = 1.8 xx 10^(-5)`,
`1.8 = 0.2553`)

A

`4.7447`

B

`3.7447`

C

`9.2553`

D

`12.523`

Text Solution

Verified by Experts

The correct Answer is:
C
`NH_(4)OH+HCl to NH_(4)Cl+_(2)O`
`NH_(4)OH]=60xx0.1-30xx0.1`
`[NH_(4)Cl]=30.01`
`:.P^(H)=14-P(kb)`
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

20 mL of 0.1 M H_2SO_4 solution is added to 30 mL of 0.2 M NH_4OH solution. The pH of the resultant mixture is : [ pK_b of NH_4OH =4.7]

20mL of 0.1M(H_(2)SO_(4)) solution is added to 30mL of 0.2M(NH_(4)OH) soulution. The ._(p)H of the resultant mixture is : [(p)k_(b) " of " NH_(4)OH=4.7] .

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))

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

10 ml of M/10 NH_(4)OH is mixed with 4ml of M/10H_(2)SO_(4) solution. The pH of the resulting solution is (pK_(b)NH_(4)OH=4.76), (log2=0.3) .

1 lit of buffer solution contains 0.1 mole each of NH_(4)OH and NH_(4)Cl . What will be the P^(H) of the solution when 0.01 mole of HCl is added to it [ P^(Kb) of NH_(4)OH = 4.74 ]

Calculate the degree of hydrolysis and pH of 0.2M solution of NH_(4)C1 Given K_(b) for NH_(4)OH is 1.8 xx 10^(-5) .

200 ml of an aqueous solution contains 0.1 M NaOH and 0.2M NH_(4)OH(aq) . In a titration experiment, 0.1 MHCI (aq) solution is slowly added to it. Given K_(b) (NH_(4)OH) = 10^(-5) On adding 400 mL of HCI solution (from start) pH of solution will be finally: