A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution

We're given that 50.0 mL of an NaOH solution is added to 100.0 mL of a 0.400 M HCl solution. Also, the solution forms 2.06 g of precipitate when treated with chromium(III) nitrate. First, let's list the relevant chemical reactions: 1. Neutralization reaction between NaOH and HCl: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) 2. Reaction between remaining OH- ions and Cr(NO3)3: 3OH^(-)(aq) + Cr(NO3)3(aq) -> Cr(OH)3(s) + 3NO3^(-)(aq) The molar mass of Cr(OH)3 is: 51.996 (Cr) + 3(15.999) (O) + 3(1.008) (H) = 151.994 g/mol Mole calculation formula for given mass: n = mass / molar mass The concentration formula: C = n / V

To determine the number of moles of Cr(OH)3 precipitate formed, we need to divide the mass of Cr(OH)3 by its molar mass: n(Cr(OH)3) = (mass of Cr(OH)3) / (molar mass of Cr(OH)3) n(Cr(OH)3) = 2.06 g / 151.994 g/mol = 0.01355 mol From the stoichiometry of the reaction between OH- ions and Cr(NO3)3, 1 unit of Cr(OH)3 precipitate is formed by reacting 3 units of OH- ions. So, we'll multiply the moles of precipitate by 3 to find moles of OH- ions. n(OH-) = 3 * n(Cr(OH)3) n(OH-) = 3 * 0.01355 mol = 0.04065 mol

Moles of HCl = (volume of HCl solution) * (concentration of HCl solution) n(HCl) = 100.0 mL * 0.400 mol/L = 0.0400 mol (considering volume in L, 100.0 mL = 0.100 L) Initially, the moles of NaOH needed to neutralize the given moles of HCl will be equal to the moles of HCl (as in the neutralization reaction, one mole of NaOH reacts with one mole of HCl). Subtracting these moles from the total moles of OH- ions will give us the moles of NaOH in 50.0 mL. n(NaOH) = n(OH-) - n(HCl) n(NaOH) = 0.04065 mol - 0.0400 mol = 0.00065 mol Now, let's find the concentration of the NaOH solution using the moles of NaOH and volume of NaOH solution: C(NaOH) = n(NaOH) / V(NaOH) C(NaOH) = 0.00065 mol / 0.050 L = 0.0130 M So, the concentration of the NaOH solution is 0.0130 M.