What is the pH of the resulting solution when | vedclass

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

Question

One mole of $NaOH$ is completely neutralised by one mole of $HCL$.

Hence, $0.01$ mole of $NaOH$ will be completely neutralised by $0.01$ mole of $H

Vedclass · Accepted Answer

$12.65$

Vedclass · Answer

One mole of $NaOH$ is completely neutralised by one mole of $HCL$.

Hence, $0.01$ mole of $NaOH$ will be completely neutralised by $0.01$ mole of $HCL$.

$\Rightarrow\; NaOH$ left unneutralised $=0.1-0.01=0.09$ mol As equal volumes of two solutions are mixed,

$[\mathrm{OH}]^{-}=\frac{0.09}{2}=0.045 \mathrm{M} $

$ \Rightarrow \mathrm{pOH}=-\log (0.045)=1.35 $

$	herefore \mathrm{pH}=14-1.35=12.65$

What is the $pH$ of the resulting solution when equal volumes of $0.1\, M\, NaOH$ and $0.01\, M\, HCl$ are mixed?

Similar Questions

A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$

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

Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)

The first and second dissociation constants of an acid $H_2A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

The $ pH$ of $ 0.1$ $M$ acetic acid is $3$, the dissociation constant of acid will be