Equal volumes of three acid solutions of pH , | vedclass

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Question

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

-or $\quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}\right]$ of solution $

Vedclass · Accepted Answer

$3.7$

Vedclass · Answer

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]$

-or $\quad\left[\mathrm{H}^{+}ight]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}ight]$ of solution $1=10^{-3}$

$\left[\mathrm{H}^{+}ight]$ of solution $2=10^{-4} ;\left[\mathrm{H}^{+}ight]$ of solution $3=10^{-5}$

Total concentration of $\left[\mathrm{H}^{+}ight]$

$=10^{-3}\left(1+1 	imes 10^{-1}+1 	imes 10^{-2}ight)$

$\Rightarrow 10^{-3}\left(\frac{1}{1}+\frac{1}{10}+\frac{1}{100}ight)$

$\Rightarrow 10^{-3}\left(\frac{100+10+1}{100}ight)$

$\Longrightarrow 10^{-3}\left(\frac{111}{100}ight)=1.11 	imes 10^{-3}$

$So, \mathrm{H}^{+}$ ion concentration in mixture of equal volume of these acid solution $=1.11 	imes 10^{-3} / 3=3.7 	imes 10^{-4} \,\mathrm{M}$

Equal volumes of three acid solutions of $pH \,3, 4$ and $5$ are mixed in a vessel. .........$ \times 10^{-4} \,M$ will be the $H^+$ ion concentration in the mixture ?

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