5drops of 0.15 m ki added to
40 drops of na2s2o3
what is the final concentration of ki?

Explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

Calculations:

(a) Calculate the total volume

V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

the answer would be a