Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
Answer:
True
Explanation:
Well,I hope it helps....
Just correct me if I'm wrong..
The concentration of OH⁻ is first converted to pOH bu using followinf formula,
pOH = -log [OH⁻]
Putting value,
pOH = -log (0.006)
pOH = 2.221
As we know,
pH + pOH = 14
Solving for pH,
pH = 14 - pOH
Putting value of pOH,
pH = 14 - 2.221
pH = 11.779
Result:
Option-1 is the correct answer.