$K_b$ for $NH_4OH$ is $1.8\times 10^{-5}.$ The $[\mathop O\limits^\Theta H]$ of $0.1\,M\,NH_4OH$ is
$K_b$ for $NH_4OH$ is $1.8\times 10^{-5}.$ The $[\mathop O\limits^\Theta H]$ of $0.1\,M\,NH_4OH$ is
- A
$5.0\times 10^{-2}$
- B
$4.20\times 10^{-3}$
- C
$1.34\times 10^{-3}$
- D
$1.8\times 10^{-6}$
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