Given that the dissociation constant for H2O is Kw =1× 10-14 what is the pH of a M KOH solution?

Question

Given that the dissociation constant for H2O is Kw =1× 10-14 what is the pH of a M KOH solution? (A)  11  (B)  3  (C)  14  (D)  10-11

Tardigrade Admin · Accepted Answer

Given, Kw=1 × 10-14 mol 2 L -2 Molarity of KOH = normality of K O H N=1 × 10-3 or [O H-]=1 × 10-3[H+] =(1 × 10-14/[O H-])=(1 × 10-14/1 × 10-3)=1 × 10-11 p H=- log [H+] p H=- log 10-11 p H=11

Q. Given that the dissociation constant for $H_{2} O$ is $K_{w} = 1 \times 1 0^{- 14}$ what is the pH of a $M K O H$ solution?

$11$

$3$

$14$

$1 0^{- 11}$

Q. Given that the dissociation constant for H2​O is Kw​=1×10−14 what is the pH of a MKOH solution?

11

3

14

10−11

Given, Kw​=1×10−14mol2L−2 Molarity of KOH= normality of KOH N=1×10−3 or [OH−]=1×10−3[H+] =[OH−]1×10−14​=1×10−31×10−14​=1×10−11 pH=−log[H+] pH=−log10−11pH=11

Q. Given that the dissociation constant for $H_{2} O$ is $K_{w} = 1 \times 1 0^{- 14}$ what is the pH of a $M K O H$ solution?

$11$

$3$

$14$

$1 0^{- 11}$