Equal volumes of three acid solutions of pH , | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 ?

Similar Questions

Calculate $pH$ of solution of $6.0$ $gm$ acetic acid in $250$ $mL$. ( ${K_a} = 1.8 \times {10^{ - 5}}$ at $298$ $K$ ) ( $C = 12, H = 1, O = 16$ )

The solution of $N{a_2}C{O_3}$ has $pH$

It has been found that the $pH$ of a $0.01$ $M$ solution of an organic acid is $4.15 .$ Calculate the concentration of the anion, the ionization constant of the acid and its $p{K_a}$

$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-

If $pK_a =\, -\,log K_a=4$ for a weak acid $HX$ and $K_a= C\alpha ^2$ then Van't Haff factor when $C = 0.01\,M$ is