A student used a cation exchange resin to determine themolarity of sodium ion in

Question

A student used a cation exchange resin to determine themolarity of sodium ion in an unknown sodium chloride solution. Thestudent diluted the 25.00mL aliquot of the original unknownsolution to 250.0mL in a volumetric flask. After passing a 50.00mLaliquot of the dilute unknown sodium chloride solution through acation exchange resin, 30.00mL of 0.010M sodium hydroxide solutionwere required to titrate the eluate a) Calculate the molarity of sodium ion in the 50.00mL portionof the diluted unknown sodium chloride solution b) Calculate the molarity of sodium ion in the original sodiumchloride solution A student used a cation exchange resin to determine themolarity of sodium ion in an unknown sodium chloride solution. Thestudent diluted the 25.00mL aliquot of the original unknownsolution to 250.0mL in a volumetric flask. After passing a 50.00mLaliquot of the dilute unknown sodium chloride solution through acation exchange resin, 30.00mL of 0.010M sodium hydroxide solutionwere required to titrate the eluate a) Calculate the molarity of sodium ion in the 50.00mL portionof the diluted unknown sodium chloride solution b) Calculate the molarity of sodium ion in the original sodiumchloride solution

Explanation / Answer

Let's start with what we know: .010M NaOH x .300L = .0030 moles of Na+ andOH- ions. Because this titrated the sodium chloride solution, we know thatthe 250mL NaCl solution has .0030 moles. Divide the moles of the NaCl solution by the Liters of solution(.250L). b) We know the initial volume, final volume, and finalconcentration of the NaCl solution, so we can use the formulaM1V1=M2V2 and solve forM1 to find the initial concentration.

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