Ionic product of water at $310 \,K$ is $2.7 \times 10^{-14}$. What is the $\mathrm{pH}$ of neutral water at this temperature?

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Ionic product,  $K_{w}=\left[ H ^{+}\right]\left[ OH ^{-}\right]$

Let $\left[ H ^{+}\right]=x$

Since $\left[ H ^{+}\right]=\left[ OH ^{-}\right], K_{ w }=x^{2}$

$\Rightarrow K_{ w }$ at $310 \,K$ is $2.7 \times 10^{-14}$.

$\therefore 2.7 \times 10^{-14}=x^{2}$

$\Rightarrow x=1.64 \times 10^{-7}$

$\Rightarrow\left[ H ^{+}\right]=1.64 \times 10^{-7}$

$\Rightarrow pH =-\log \left[ H ^{+}\right]$

$=-\log \left[1.64 \times 10^{-7}\right]$

$=6.78$

Hence, the $pH$ of neutral water is $6.78$

Similar Questions

Accumulation of lactic acid $(HC_3H_5O_3),$ a monobasic acid in tissues leads to pain and a feeling of fatigue. In a $0.10\, M$ aqueous solution, lactic acid is $3.7\%$ dissociates. The value of dissociation constant, $K_a,$ for this acid will be

  • [NEET 2013]

Derive the equation of ionization constants ${K_a}$ of weak acids $HX$.

The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05\, M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.

The molar conductivity of a solution of a weak acid $HX (0.01\ M )$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $HY (0.10 \ M )$. If $\lambda_{ X }^0 \approx \lambda_{ Y ^{-}}^0$, the difference in their $pK _{ a }$ values, $pK _{ a }( HX )- pK _{ a }( HY )$, is (consider degree of ionization of both acids to be $\ll 1$ )

  • [IIT 2015]

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