Dissociation constant for a monobasic acid is | vedclass

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Question

$\mathrm{K}_{\mathrm{a}}=10^{-4}=\mathrm{C}_{0} \alpha^{2} \quad \% \alpha=2 \%$

$\alpha=\frac{2}{100}=0.02$

$\because C_{0}=\frac{K_{a}}{\alpha

Vedclass · Accepted Answer

$2.3$

Vedclass · Answer

$\mathrm{K}_{\mathrm{a}}=10^{-4}=\mathrm{C}_{0} \alpha^{2} \quad \% \alpha=2 \%$

$\alpha=\frac{2}{100}=0.02$

$\because C_{0}=\frac{K_{a}}{\alpha^{2}}=\frac{10^{-4}}{0.02 	imes 0.02}=\frac{1}{4}$

$\left[\mathrm{H}^{+}ight]=\mathrm{C}_{\mathrm{o}} \alpha$

$=\frac{1}{4} 	imes 0.02=0.005 \Rightarrow 5 	imes 10^{-3}$

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]$

$=-\log \left[5 	imes 10^{-3}ight]$

$=-[\log 5-3 \log 10]$

$=2.3$

Base	$K _{ b }$
Dimethylamine, $\left( CH _{3}\right)_{2} NH$	$5.4 \times 10^{-4}$
Triethylamine, $\left( C _{2} H _{5}\right)_{3} N$	$6.45 \times 10^{-5}$
Ammonia, $NH _{3}$ or $NH _{4} OH$	$1.77 \times 10^{-5}$
Quinine, ( $A$ plant product)	$1.10 \times 10^{-6}$
Pyridine, $C _{5} H _{5} N$	$1.77 \times 10^{-9}$
Aniline, $C _{6} H _{5} NH _{2}$	$4.27 \times 10^{-10}$
Urea, $CO \left( NH _{2}\right)_{2}$	$1.3 \times 10^{-14}$

Dissociation constant for a monobasic acid is $10^{-4}$ . What is the $pH$ of the monobasic acid ? (If $\%$ dissociation $= 2\,\%$ )

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