Ph of 0.005M BaoH2 is

1.	Ph of 0.005M BaoH2 is
Answer» Answer: CONCENTRATION of OH − in the SOLUTION = 0.005×2=0.01mole/l pOH=−log[0.01]=2 pH=14−pOH=14−2=12 When 100 ml of this solution DILUTED to 1000 ml Moles of OH − =0.01 mole Concentration of OH − = 1000 0.01×100 =0.001 pOH=−log[10 −3 ]=3 pH=14−3=11 Explanation:

Answer»

Answer:

CONCENTRATION of OH

−

in the SOLUTION = 0.005×2=0.01mole/l

pOH=−log[0.01]=2

pH=14−pOH=14−2=12

When 100 ml of this solution DILUTED to 1000 ml

Moles of OH

−

=0.01 mole

Concentration of OH

−

1000

0.01×100

=0.001

pOH=−log[10

−3

]=3

pH=14−3=11

Explanation:

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