Please help: If 542 microLiters of 0.0402 M KMnO4 are required to titrate 200 microLiters of NaI solution...?

Question

...to the endpoint in acid, what was the concentration of Iodide ion in the original solution?

Any help will be appreciated! Thank you.

Answer 1

M1V1=M2V2

200*(?) = 542*0.0402

this gives you molarity of NaI - 0.108M

NaI is built of Na = 23g and I of 127 gms totalling 150 gms.

So your 0.108 molar solution has 15 gms of NaI.

now, 150 gms NaI contains 127 gms of I

than 15 gms contains = 15*127/150 = 12.7 gms I in liter.

1liter = 1000ml = 1000000 ul contains 12.7 gms I

than 200 ul contains = 200*12.7/1000000= 2.54x10^-3

ANS = 2.54x10^ -3