Science, asked by aafrin31, 1 year ago

0.002 M kOH calculate pH

Answers

Answered by Anonymous
6


(d) 0.002 M KOH

KOH ↔ K+ + OH-

[OH-] = [KOH]

[OH-] = 0.002

pOH = - log [OH-] =-log[0.002] = 2.69

pH= 14-2.69 = 11.70

aafrin31: everything is ok but the value 2.69 why you used this i didn't understand
aafrin31: ohh ... thanks
saud59: More clearly -log[2×10^-3]=3-0.3010=2.669 as log(2) is 0.3010 and using propert log(a×b)=log(a)+log(b)
saud59: hope now it is crystal clear
aafrin31: you both are my brainliest ..tysm
saud59: thanks
Answered by saud59
2

0.002 M KOH

KOH ↔ K+ + OH-

[OH-] = [KOH]

[OH-] = 0.002

pOH = - log [OH-] =-log[0.002] = 2.69

pH= 14-2.69 = 11.70

hope you understand
aafrin31: tysm i understand easily now.
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