If a chemist titrates 125 mL of sodium hydroxide with a 6.2 M solution of hydrochloric acid and the titraiton requires 65.0 mL of the acid to reach the end point. What is the concentration of the sodium hydroxide?
BALANCE equationNaOH + HCl --> NaCl + H2OCALCULATE number of moles of HCl present65.0 mL HCl x 1 L/1000 mL x 6.2 mol HCl/ 1 HCl = .031 mol HClUSE MOLAR RATIO from equation to convert from moles of HCl to moles of NaOH.031 mol HCl x 1 mol NaOH/1 mol HCl = .031 mol NaOHCALCULATE concentrationConcentration = number of moles/volumeVolume of NaOH = 125 mL = .125 L.031 mol NaOH/.125 L NaOHConcentration = .248 mol/L = .248 M
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