The kw of pure water at 40â°c is 2.92 x 10^-14.
(a) calculate the [h+] and [oh -] in pure water at 40â°c.
(b) what is the ph of pure water at 40â°c?
(c) if the hydroxide ion concentration in a solution is 0.10 m, what is the ph at 40â°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

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