Answer:Sample Absorbance (625 nm)
A 0.536
B 0.783
C 0.045
Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.
First, we see 1 mole of NH3 gives 1 mole product.
In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA
( mA = moles of NH3 in A) vol of B = 25 = vol of A
now A = el C = eC ( since l = 1cm)
Because, n net absorbance due to complex blank absorbance must be removed.
Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738
(you can plug in different numbers in this step)
A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491
So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)
Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M
Lake water vol = 10 ml out of 25,
Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M
Then, A = 0.491 = e x 1 x 1.093 x10^-4
e = 4492 M-1cm-1
Explanation:
