Question

What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233 O E) 8.446

Answer 1

Answer:

PH= 6.767     (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

$[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2}$

at 40ºC

$[H_{3}O^{+} ] = [OH^{-} ]$

$[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}$

$[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M$

$PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767$