hsc chem question help (1 Viewer)

jimmysmith560 · Jun 30, 2022

Would the following working help?

$\left[H_3O^+\right]_{eq}=10^{-4.46}\:\text{mol L}^{-1}$

	$\left(CH_3\right)_3NH^+$	$H_3O^+$	$\left(CH_3\right)_3N$
Initial	N/A	0	0
Change	N/A	$+10^{-4.46}$	$+10^{-4.46}$
Equilibrium	$x$	$10^{-4.46}$	$10^{-4.46}$

$K_a=\frac{\left[H_3O^+\right]\left[\left(CH_3\right)_3N\right]}{\left[\left(CH_3\right)_3NH^+\right]}$

$1.55\times 10^{-10}=\frac{10^{-4.46}\times 10^{-4.46}}{x}$

$\rightarrow \:x=7.75\:\text{mol L}^{-1}$

$=\left[\left(CH_3\right)_3NH^+\right]_{sat}$

$\left(CH_3\right)_3NHCl_{\left(s\right)}$ ⇌ $\left(CH_3\right)_3NH^+\:_{\left(aq\right)}+Cl^-\:_{\left(aq\right)}$

$\left[Cl^-\right]_{sat}=7.75\:\text{mol L}^{-1}$

$K_{sp}=\left[\left(CH_3\right)_3NH^+\right]\left[Cl^-\right]$

$=7.75\times 7.75$

$=60.2\:\text{i.e. (C)}$

Run hard@thehsc · Jun 30, 2022

phew got that

hsc chem question help (1 Viewer)

astraaaaaah

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jimmysmith560

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Run hard@thehsc

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