Question

What quantity of chloride ion in water supply was determined by the titrating the sample against AgNO3 solution

Answer 1

Solution.

i)

Since according to the reaction equation, the number of moles of silver nitrate is equal to the number of moles of chloride ions, we find the number of moles of chloride ions:

$n(AgNO3)=n(Cl^-) = 0.1 \times 0.0202=0.00202 \ mol$

$m(Cl^-) = 0.00202 \times 35.457 = 0.07 \ g$

$w(Cl^-) = \frac{0.07}{10} \times 100 \% = 0.7 \ \%$

ii)

w(Cl-) = 7000 ppm

Answer:

i)

w(Cl-) = 0.7 %

ii)

w(Cl-) = 7000 ppm