169. 17 a student titrates a 25.00 ml sample of well water with 18.2 ml 0.100 m agno₃ to completely…

169. 17 a student titrates a 25.00 ml sample of well water with 18.2 ml 0.100 m agno₃ to completely precipitate the chloride ion. the cl⁻ is a. 1.82×10⁻³ m b. 7.28×10⁻² m c. 1.37×10⁻¹ m d. 1.50×10⁻¹ m

169. 17 a student titrates a 25.00 ml sample of well water with 18.2 ml 0.100 m agno₃ to completely precipitate the chloride ion. the cl⁻ is a. 1.82×10⁻³ m b. 7.28×10⁻² m c. 1.37×10⁻¹ m d. 1.50×10⁻¹ m

Answer

Explanation:

Step1: Write the chemical reaction

The reaction between $AgNO_3$ and $Cl^-$ is $AgNO_3(aq)+Cl^-(aq)=AgCl(s)+NO_3^-(aq)$. The mole - ratio of $AgNO_3$ to $Cl^-$ is 1:1.

Step2: Calculate the moles of $AgNO_3$

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters. $V_{AgNO_3}=18.2\ mL = 0.0182\ L$ and $M_{AgNO_3}=0.100\ M$. So $n_{AgNO_3}=M_{AgNO_3}\times V_{AgNO_3}=0.100\ mol/L\times0.0182\ L = 0.00182\ mol$.

Step3: Determine the moles of $Cl^-$

Since the mole - ratio of $AgNO_3$ to $Cl^-$ is 1:1, $n_{Cl^-}=n_{AgNO_3}=0.00182\ mol$.

Step4: Calculate the concentration of $Cl^-$

$V_{sample}=25.00\ mL = 0.02500\ L$. Use the formula $M=\frac{n}{V}$, so $M_{Cl^-}=\frac{n_{Cl^-}}{V_{sample}}=\frac{0.00182\ mol}{0.02500\ L}=0.0728\ M = 7.28\times 10^{-2}\ M$.

Answer:

B. $7.28\times 10^{-2}\ M$