Question

the concentration of sn²⁺ in a solution is determined by titrating it with a 0.1278 m ce⁴⁺ solution. the balanced net - ionic equation for the reaction is: 2ce⁴⁺(aq) + sn²⁺(aq) → 2ce³⁺(aq) + sn⁴⁺(aq). in one experiment, 16.05 ml of the 0.1278 - m ce⁴⁺ solution is required to react completely with 20.00 ml of the sn²⁺ solution. calculate the concentration of the sn²⁺ solution.

Explanation:

Step1: Determine the mole - ratio from the balanced equation

The balanced net - ionic equation is \(2Ce^{4 +}(aq)+Sn^{2+}(aq)
ightarrow2Ce^{3 +}(aq)+Sn^{4+}(aq)\). The mole - ratio of \(Ce^{4+}\) to \(Sn^{2+}\) is \(n_{Ce^{4+}}:n_{Sn^{2+}} = 2:1\).

Step2: Calculate the moles of \(Ce^{4+}\) used

Use the formula \(n = M\times V\), where \(M\) is the molarity and \(V\) is the volume in liters. The volume of \(Ce^{4+}\) solution \(V_{Ce^{4+}}=16.05\ mL = 0.01605\ L\) and molarity \(M_{Ce^{4+}} = 0.1278\ M\). So, \(n_{Ce^{4+}}=M_{Ce^{4+}}\times V_{Ce^{4+}}=0.1278\ mol/L\times0.01605\ L = 0.00205119\ mol\).

Step3: Calculate the moles of \(Sn^{2+}\)

Since \(n_{Ce^{4+}}:n_{Sn^{2+}} = 2:1\), then \(n_{Sn^{2+}}=\frac{n_{Ce^{4+}}}{2}\). So, \(n_{Sn^{2+}}=\frac{0.00205119\ mol}{2}=0.001025595\ mol\).

Step4: Calculate the molarity of \(Sn^{2+}\)

The volume of \(Sn^{2+}\) solution \(V_{Sn^{2+}} = 20.00\ mL=0.02000\ L\). Using the formula \(M=\frac{n}{V}\), we have \(M_{Sn^{2+}}=\frac{n_{Sn^{2+}}}{V_{Sn^{2+}}}=\frac{0.001025595\ mol}{0.02000\ L}=0.05128\ M\).

Answer:

\(0.05128\)