Question

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5.6g of KOH (mol. wt = 56) is present in 1 litre of solution. It's pH value is
a) 1
b) 13
c) 14
d) 0​

Answer 1

Answer:

The answers is ph value is will be 1

Answer 2

Answer:

KOH → k+ + oh-

t=0 5.6/56=0.1mole 0 0

t=t 0.1 0.1

[since koh is strong base so dissociate completely]

[ oh-]= 0.1/1000=1×10^-4

Poh = -log 10^-4

= 4

ph = 14-4=10