Answer to Question #155888 in Inorganic Chemistry for Holly

Question

I take 5g of solid-liquid mixture sample,after acid digestion and make up to 50ml,from which I take 10 ml for chloride analysis,there 0.2ml AgNO3 consumed after back titration against 0.02N AgNO3 and NH4SCN.

how to calculate the chloride in the original sample? unit of result as mg/L or mg/kg?

Answer 1

from,

molarity × number of hydrogen ions = normality

molarity × 1 = 0.02N

Therefore, molarity of AgNO₃ = 0.02M

$AgNO_3 + Cl^- \to AgCl + NO_3^-$

from, $\dfrac{c_a×v_a}{c_b×v_b} = \dfrac{n_a}{n_b}$

$\dfrac{c_a×10ml}{0.02×0.2ml} = \dfrac{1}1$

$c_a = \dfrac{0.02×0.2}{10}$ = 0.0004M

molar mass of HCl = 36.5g/mol

molar concentration = 0.0004M

mass concentration = molar mass × molar concentration = 36.5 × 0.0004 = 0.0146 g/L = 14.6mg/L

Therefore, the concentration of the chloride in the original sample is 14.6mg/L.