Question

question 6
13 pts
to determine the concentration of chloride ion (cl⁻) in a 100. ml sample of ground water, a chemist adds a large enough volume of agno₃ solution to precipitate all cl⁻ as agcl. the mass of the resulting precipitate is 96.0 mg. what is the chloride ion concentration in milligrams of chloride per liter of ground water?

Explanation:

Step1: Find molar mass of AgCl

The molar mass of AgCl: $M_{AgCl}=107.87\ g/mol + 35.45\ g/mol=143.32\ g/mol$.

Step2: Calculate moles of AgCl

$n_{AgCl}=\frac{m_{AgCl}}{M_{AgCl}}$, given $m_{AgCl}=96.0\ mg = 0.096\ g$. So $n_{AgCl}=\frac{0.096\ g}{143.32\ g/mol}\approx6.70\times10^{-4}\ mol$.

Step3: Determine moles of Cl⁻

From the chemical reaction $Ag^{+}+Cl^{-}
ightarrow AgCl\downarrow$, the mole - ratio of $Cl^{-}$ to $AgCl$ is 1:1. So $n_{Cl^{-}} = n_{AgCl}=6.70\times10^{-4}\ mol$.

Step4: Calculate mass of Cl⁻

The molar mass of $Cl^{-}$ is $M_{Cl^{-}} = 35.45\ g/mol$. So $m_{Cl^{-}}=n_{Cl^{-}}\times M_{Cl^{-}}=6.70\times10^{-4}\ mol\times35.45\ g/mol\approx0.0238\ g = 23.8\ mg$.

Step5: Calculate concentration of Cl⁻

The volume of the sample $V = 100\ mL=0.1\ L$. The concentration $C=\frac{m_{Cl^{-}}}{V}$. So $C=\frac{23.8\ mg}{0.1\ L}=238\ mg/L$.

Answer:

238 mg/L