What molar concentration of NHprovides a[OH-] of 1.5x10-3? (Kb=1.8x10-5)

  • Option 1)

    0.125m

  • Option 2)

    (0.125+1.5x10-3)

  • Option 3)

    (0.124-1.5x10-3)m

  • Option 4)

    (1.5x10-3)m

 

Answers (1)
V Vakul

As we discussed in concept

Value of Kb -

MOH(aq)\rightleftharpoons M^{+}(aq)+\bar{O}H(aq)

- wherein

K_{b}=\frac{[M^{+}]\:[\bar{O}H]}{[MOH]}


K_{b}=\frac{C\alpha ^{2}}{1-\alpha }

C=initial\:concentration\:of\:base

\alpha =degree\:of\:ionization\:of\:base

 

 The reaction goes as follows

 NH_{3}\:(aq)\ {+}\:\:H_{2}O{(l)}\rightleftharpoons \:NH_{4}^{+}{(aq)}+OH^{-}{(aq)}

K_{b}=\frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}

We know that [NH_{4}^{+}]=[OH^{-}]=1.5\times 10^{-3}

=>\:\:\:\:1.8\times 10^{-5}=\frac{(1.5\times 10^{-3})^{2}}{[NH_{3}]}

=>\:\:\:\:[NH_{3}]=0.125\:M


Option 1)

0.125m

This option is correct.

Option 2)

(0.125+1.5x10-3)

This option is incorrect.

Option 3)

(0.124-1.5x10-3)m

This option is incorrect.

Option 4)

(1.5x10-3)m

This option is incorrect.

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