2. The mass percent of chloride ion in a seawater sample is determined by titrat

Question

2. The mass percent of chloride ion in a seawater sample is determined by titrating 25.00 mL of seawater with silver nitrate solution causing the precipitation of AgCl. The end point of the titration is reached when all of the chloride anion has reacted and only free silver ions are present. If 53.63 mL of 0.2970 M AgNO3 is required to reach the end point, what is the mass percent of Cl- in the seawater (Density of seawater = 1.024 g/mL)?

2. The mass percent of chloride ion in a seawater sample is determined by ti seawater with silver nitrate reached when all of the chloride anion has reacted and only free silver ions are p 0.2970 M AgNOs is required to reach the end point, what is the mass percent of Cr in the seawater (Density of seawater 1.024 g/mL)? trating 25.00 mL of solution causing the precipitation of AgCl. The end point of the titration is resent. If 53.63 mL of Net Ionic equation: Ag+(aq) + Cl(aq) AgCl(s)

Explanation / Answer

Solution:- By balancing the equation

Cl-(aq) + AgNO3(aq) ---> AgCl(s) + NO3-(aq)

notice the 1:1 mole ratio of Cl- to AgCl(s)

then... convert mL AgNO3 to g Cl-...
53.63mL AgNO3 x (1L / 1000mL) x (0.2970moles AgNO3 / L) x (1 mole Cl- / 1 mole AgNO3) x (35.45g Cl- / mole Cl-) = 0.56465g Cl- (per 25.00 mL solution!)

then...calculating %...
(0.56465g Cl / 25.00mL solution) x (1mL solution / 1.024g solution) = 0.0220 = 2.2%

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