1.38 g of sodium hydrogensulfite (NaHSO3; molar mass = 104.061 g.mol-1) was dissolved in 100.0 mL water. What is the pH of the resultant solution, given the Ka of HSO3− = 6.60 x 10−8?

Sure, here are the steps to solve this problem:

Step 1: Calculate the molarity of NaHSO3 The molarity (M) is calculated by dividing the mass of the solute by its molar mass and then dividing by the volume of the solution in liters.

M = (1.38 g / 104.061 g/mol) / 0.100 L = 0.1326 M

Step 2: Write the ionization equation for HSO3- HSO3- (aq) ↔ H+ (aq) + SO3 2- (aq)

Step 3: Set up the ICE table The ICE table stands for Initial, Change, and Equilibrium. It's a way to organize the information for the calculation.

Initial: [HSO3-] = 0.1326 M, [H+] = 0, [SO3 2-] = 0 Change: [HSO3-] = -x, [H+] = +x, [SO3 2-] = +x Equilibrium: [HSO3-] = 0.1326 - x, [H+] = x, [SO3 2-] = x

Step 4: Write the expression for Ka and substitute the equilibrium concentrations Ka = [H+][SO3 2-] / [HSO3-] = x * x / (0.1326 - x) = 6.60 x 10^-8

Step 5: Solve for x This is a quadratic equation, but if we assume that x is much smaller than 0.1326 (which is a valid assumption because Ka is very small), we can simplify the equation to x^2 / 0.1326 = 6.60 x 10^-8. Solving for x gives x = sqrt(0.1326 * 6.60 x 10^-8) = 7.53 x 10^-5 M.

Step 6: Calculate the pH The pH is calculated as -log[H+], so the pH of the solution is -log(7.53 x 10^-5) = 4.12.