Question

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?

Answer 1

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

b)pH = 6.76

c)pH=13

Step-by-step 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