The Kw of pure water at 40°C is 2.92 x 10^-14.

(a) Calculate the [H+] and [OH - ] in pure water at 40°C.

(b) What is the pH of pure water at 40°C?

(c) If the hydroxide ion concentration in a solution is 0.10 M, what is the pH at 40°C?

a) [H+]=[OH-] = 1.708e-7

b) pH = 6.76

c) pH=13

Explanation:

a) Kw=[H+] (OH-) = 2.92e-14

As [H+]=[OH-]=take it equal to x

x2=2.92e-14

take squre root on both sides

we get x=1.708e-7=[H+]=[OH-]

b) As we know that pH = - log[H+]

putting the value in above equaton we get

pH = - log[1.708e-7]

pH = 6.76

c) if [OH-]=0.10 M FIND pH

pOH = - log (OH-)

pOH = - log (1e-1)

pOH = 1 but we also know that pH+pOH=14

sp pH=14-pOH

pH=14-1

pH=13