- Home
- Standard 11
- Chemistry
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 ?
$37$
$11.1$
$1.11$
$3.7$
Solution
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$
-or $\quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}\right]$ of solution $1=10^{-3}$
$\left[\mathrm{H}^{+}\right]$ of solution $2=10^{-4} ;\left[\mathrm{H}^{+}\right]$ of solution $3=10^{-5}$
Total concentration of $\left[\mathrm{H}^{+}\right]$
$=10^{-3}\left(1+1 \times 10^{-1}+1 \times 10^{-2}\right)$
$\Rightarrow 10^{-3}\left(\frac{1}{1}+\frac{1}{10}+\frac{1}{100}\right)$
$\Rightarrow 10^{-3}\left(\frac{100+10+1}{100}\right)$
$\Longrightarrow 10^{-3}\left(\frac{111}{100}\right)=1.11 \times 10^{-3}$
$So, \mathrm{H}^{+}$ ion concentration in mixture of equal volume of these acid solution $=1.11 \times 10^{-3} / 3=3.7 \times 10^{-4} \,\mathrm{M}$
Similar Questions
What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from Table . Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
| 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, $CO \left( NH _{2}\right)_{2}$ | $1.3 \times 10^{-14}$ |
