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Question

Chemistry Question on p -Block Elements

The pOH of 0.0005 M sulphuric acid is

A

5

B

3

C

11

D

12

Answer

11

Explanation

Solution

H2SO4<=>2H++SO42− { H_2SO_4 <=> 2H^+ +SO_4^{2-}}
So, [H+]=2×0.0005=1×10−3 M {[H^+] = 2\times 0.0005 = 1 \times 10^{-3} \, M }
pH=−log[H+]=−log(1×10−3){pH = - log [H^+] = - log(1\times 10^{-3}) }
pH=3{pH = 3 }
∵pH+pOH=14\because {pH +pOH = 14 }
???3+pOH=14 {3 +pOH = 14 }
????????pOH=14−3 { pOH = 14 - 3 }
????????pOH=14 {pOH = 14 }
????????pOH=11 {pOH = 11 }