The ionic product of water at 50^{circ} C is | vedclass

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

Question

(C) The ionic product of water is given by $K_w = [H^+][OH^-]$.For pure water,$[H^+] = [OH^-]$,so $K_w = [H^+]^2$.Given $K_w = 4 	imes 10^{-14}$ at $

Vedclass · Accepted Answer

$6.7$ and neutral

Vedclass · Answer

(C) The ionic product of water is given by $K_w = [H^+][OH^-]$.For pure water,$[H^+] = [OH^-]$,so $K_w = [H^+]^2$.Given $K_w = 4 	imes 10^{-14}$ at $50^{\circ} C$.Therefore,$[H^+] = \sqrt{4 	imes 10^{-14}} = 2 	imes 10^{-7} \ M$.The $pH$ is calculated as $pH = -\log[H^+] = -\log(2 	imes 10^{-7}) = -(\log 2 + \log 10^{-7}) = -(0.301 - 7) = 6.699 \approx 6.7$.Since the concentration of $H^+$ ions is equal to the concentration of $OH^-$ ions in pure water,the water remains neutral despite the $pH$ value being less than $7$.

The ionic product of water at $50^{\circ} C$ is $4 \times 10^{-14}$. The $pH$ and nature of water is :

Similar Questions

Calculate the hydrogen ion concentration in the following biological fluids whose $pH$ are given below:
$(a)$ Human muscle-fluid,$6.83$
$(b)$ Human stomach fluid,$1.2$
$(c)$ Human blood,$7.38$
$(d)$ Human saliva,$6.4$

What is the difference between the terms 'hydrolysis' and 'hydration'?

Explain the $pH$ scale and $pH$.

Calculate the $[OH^{-}]$ if $pOH$ of a solution is $4.94$.

Calculate the $pOH$ of a solution at $25 \, ^\circ C$ that contains $1 \times 10^{-10} \, M$ of hydronium ions,i.e.,$H_3O^{+}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module