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

PH= 6.767 (answer is the A option)

Step-by-step 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`