At a certain temperature,if pK_w = 12.60,then | vedclass

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

Question

(C) Given $pK_w = 12.60$,so $K_w = [H^{+}][OH^{-}] = 10^{-12.60}$.For a neutral solution,$[H^{+}] = [OH^{-}]$.Thus,$[H^{+}]^2 = 10^{-12.60} = 10^{0.40

Vedclass · Accepted Answer

$[OH^{-}] < 5 	imes 10^{-7} \ M$

Vedclass · Answer

(C) Given $pK_w = 12.60$,so $K_w = [H^{+}][OH^{-}] = 10^{-12.60}$.For a neutral solution,$[H^{+}] = [OH^{-}]$.Thus,$[H^{+}]^2 = 10^{-12.60} = 10^{0.40} 	imes 10^{-13} \approx 2.51 	imes 10^{-13} \approx 0.251 	imes 10^{-12}$.Calculating $[H^{+}] = \sqrt{10^{-12.60}} = 10^{-6.30} \approx 5.01 	imes 10^{-7} \ M$.For an acidic solution,$[H^{+}] > 5.01 	imes 10^{-7} \ M$.Since $[H^{+}][OH^{-}] = 10^{-12.60}$,if $[H^{+}] > 5.01 	imes 10^{-7} \ M$,then $[OH^{-}] < 5.01 	imes 10^{-7} \ M$.

At a certain temperature,if $pK_w = 12.60$,then which of the following is correct for an acidic solution?

Similar Questions

The $pH$ of pure water or neutral solution at $50\,^{\circ}C$ is .... ($pK_w = 13.26$ at $50\,^{\circ}C$)

The $pH$ of a solution is $9.5$. The solution is

The $pH$ of a solution is increased from $3$ to $6$. Its ${H^{+}}$ ion concentration will be

At $25 \, ^\circ C$,the ionic product of water $(K_w)$ is equal to .......

The $pH$ of a solution is $2$. If its $pH$ is to be raised to $4$,then the $[H^{+}]$ of the original solution has to be

Vedclass Test Series

Exam Paper Generator

Online Exam Module