Home
Class 12
CHEMISTRY
How many moles of NH(3) must be added to...

How many moles of `NH_(3)` must be added to `1.0L` of `0.75M AgNO_(3)` in order to reduce the `[Ag^(o+)]` to `5.0 xx 10^(-8)M. K_(f) Ag (NH_(3))_(2)^(o+) = 1 xx 10^(8)`.

Text Solution

Verified by Experts

The correct Answer is:
1.89
`Ag^(+) + 2NH_(3) to [Ag(NH_(3))_(2)]^(+)`
`K_(f)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=1xx10^(8)`
`[AgNO_(3)]=[Ag^(+)]_("initially")=0.750`
`1xx10^(8)=0.750/(5xx10^(-8)xx[NH_(3)]^(2))`
`[NH_(3)]^(2)=0.15`
`[NH_(3)]`in solution at equilibrium=0.39M
`[NH_(3)]` added=0.39 + `[NH_(3)]` needed to convert 0.750 M `Ag^(+)` to [Ag(NH_(3))_(2)]^(+)`
`0.39+2xx0.750=1.89mol`
Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise JEE MAIN ( ARCHIVE )|50 Videos

  • IONIC EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise JEE ADVANCED( ARCHIVE )|65 Videos

  • IONIC EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise LEVEL 2|50 Videos

  • INTRODUCTION TO ORGANIC CHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ARCHIVE|81 Videos

  • JEE MAIN - 5

    VMC MODULES ENGLISH|Exercise PART II : CHEMISTRY (SECTION - 2)|5 Videos

Similar Questions

Explore conceptually related problems

Calculated the minimum amount of Nh_(3) which must be added to 1.0L of solution in order to dissolve 0.1mol AgC1 by forming [Ag(NH_(3))_(2)]^(o+)? K_(sp) of AgC1 = 1 xx 10^(-10), K_(f) (NH_(3))_(2)^(o+) = 1 xx 10^(8) .

When 0.1 mole of NH_(3) is dissolved in water to make 1.0 L of solution , the [OH^(-)] of solution is 1.34 xx 10^(-3) M. Calculate K_(b) for NH_(3) .

When 0.1 mole of NH_(3) is dissolved in water to make 1.0 L of solution, the [OH^(-)] of solution is 1.30xx10^(-3)M. Calculate K_(b) for NH_(3).

How many moles of HCI can be added to 1.0L of solution of 0.1M NH_(3) and 0.1 M NH_(4)CI without changing pOH by more than one unit? (pK_(b) of NH_(3) = 4.75)

a. Calculate [Ag^(o+)] in a solution of [Ag(NH_(3))_(2)^(o+)] prepared by adding 1.0 xx 10^(-3)mol AgNO_(3) to 1.0L of 1.0M NH_(3) solution K_(f) Ag(NH_(3))_(2)^(o+) = 10^(8) . b. Calculate [Ag^(o+)] which is in equilibrium with 0.15M [Ag(NH_(3))_(2)]^(o+) and 1.5 NH_(3) .

If 500 mL of 0.4 M AgNO_(3) is mixed with 500 mL of 2 M NH_(3) solution then what is the concentration of Ag(NH_(3))^(+) in solution? Given : K_(f1)[Ag(NH_(3))]^(+)=10^(3),K_(f2)[Ag(NH_(3))_(2)^(+)]=10^(4)

What mass of Agl will dissolve in 1.0 L of 1.0 M NH_(3) ? Neglect change in conc. Of NH_(3) . [Given: K_(sp)(AgI)=1.5xx10^(-16)), K_(f)[Ag(NH_(3))_(2)^(+)]=1.6xx10^(7)], (At. Mass Ag=108,1=127)

How many moles of NaOH can be added to 0.1L of solution of 0.1M NH_(3) and 0.1M NH_(4)CI without changing pOH by more than pne unit (pK_(a) "of" NH_(3) = 4.75) ?

For NH_(3) , K_(b)=1.8xx10^(-5) . K_(a) for NH_(4)^(+) would be

To a solution of 0.01M Mg^(2+) and 0.8M NH_(4)CI , and equal volume of NH_(3) is added which just gives precipitates. Calculate [NH_(3)] in solution. K_(sp) of Mg(OH)_(2) = 1.4 xx 10^(-11) and K_(b) of NH_(4)OH = 1.8 xx 10^(-5) .