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What will be the amount of (NH(4))(2)SO(...

What will be the amount of `(NH_(4))_(2)SO_(4)` (in g) which must be added to 500 mL of 0.2 M `NH_(4)OH` to yield a solution of pH 9.35? `["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]`

A

`5.35`

B

`6.6`

C

`1.67`

D

`10.2`

Text Solution

Verified by Experts

The correct Answer is:
B
`P^(H)=14-P^(kb)-"log"(N_(S)V_(S))/(N_(b)V_(B))` `4.74+9.26=14-logNsVs-log10^(-1)`
`14=14-log NsVs+1`
`=13=15 log NsVs`
`log NsVs=15-13=2`
`NsVs=100`
`(wt)/(66)xx(1000)/(Vs)xxVs=100`
`wt=6.6g`
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Amount of (NH_(4))_(2)SO_(4) which must be added to 50mL of 0.2 M NH_(4)OH solution to yield a solution of pH 9.26 is ( pK_(b) of NH_(4)OH=4.74 )

Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500ml of 0.2MNH_(3) to give a solution of pH=9.3 .Given pK_(b) for NH_(3)=4.7 .

Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(3) to yield a solution of pH=9 , K_(b) for NH_(3)=2xx10^(-5)

Calculate the weight of (NH_(4))_(2)SO_(4) which must be added to 500mL of 0.2M NH_(3) to yield a solution of pH = 9.35. K_(a) for NH_(3) = 1.78 xx 10^(-5) .

To prepare a buffer of pH 8.26 , amount of (NH_(4))_(2)SO_(4) to be added into 500mL of 0.01M NH_(4)OH solution [pK_(a)(NH_(4)^(+))=9.26] is:

When NH_(4)Cl is added to a solution of NH_(4)OH :

Knowledge Check

  • Amount of (NH_(4))_(2)SO_(4) which must be added to 50mL of 0.2 M NH_(4)OH solution to yield a solution of pH 9.26 is ( pK_(b) of NH_(4)OH=4.74 )

    A
    `0.10 mol`
    B
    `0.20 mol`
    C
    `0.05 mol`
    D
    `0.40 mol`

  • Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(3) to yield a solution of pH=9 , K_(b) for NH_(3)=2xx10^(-5)

    A
    `3.248 g`
    B
    `4.248 g`
    C
    `1.320 g`
    D
    `6.248 g`

  • Find the amount of (NH_(4))_(2)SO_(4) in grams which must added to 500 ml of 0.2M of NH_(4)OH to yield a solution of p^(H) = 4.74? (P^(k_(b)) = 9.26)

    A
    `5.35`
    B
    `6.6`
    C
    `1.67`
    D
    `10.2`