Question

A solution contains 0.05 M AgNO3 and 0.10 M KCl. A silver electrode is dipped in the solution and a potential of 0.53 V is measured against a standard hydrogen electrode. Calculate the concentration of Ag+ ions in the solution.

Answer 1

To calculate the concentration of Ag+ ions in the solution, we can use the Nernst equation:E = E -  RT/nF  * ln Q Where:E = measured potential  0.53 V E = standard reduction potential for Ag+/Ag  0.7996 V R = gas constant  8.314 J/molK T = temperature  assume 298 K, room temperature n = number of electrons transferred  1 for Ag+/Ag F = Faraday's constant  96485 C/mol Q = reaction quotient, which is [Ag+]/[Ag]First, let's solve for Q:0.53 V = 0.7996 V -  8.314 J/molK * 298 K /  1 * 96485 C/mol   * ln Q Now, let's isolate ln Q :ln Q  =  0.7996 V - 0.53 V  *  1 * 96485 C/mol  /  8.314 J/molK * 298 K ln Q  = 3.312Now, exponentiate both sides to solve for Q:Q = e^3.312Q = 27.45Since Q = [Ag+]/[Ag], and we know the initial concentration of AgNO3 is 0.05 M, we can set up the following equation:27.45 = [Ag+]/ 0.05 M - [Ag+] Now, solve for [Ag+]:[Ag+] = 0.05 M /  1 + 27.45 [Ag+] = 0.05 M / 28.45[Ag+]  0.00176 MSo, the concentration of Ag+ ions in the solution is approximately 0.00176 M.