Question

I'm not sure about how I'm supposed to find NaOH concentration from a strong acid and strong base titration. I'm thinking that 2.5×10^-8 mol per L that I calculated would make the moles of NaOH 2.5x10^-8 mol since it's a one to one ratio.

Hydrochloric acid titration: Confirming the concentration of NaOH Concentration of HCl solution from the stock bottle: _ 0.OSM! Initial pH of HCl solution: 1.66 pH Volume NaOH (mL) Concentration of NaOH (M) Non Phenolphthalein (end point) 11.63 13.4mL Titration curve (equivalence point) 7.60 12.5ml Show your calculation of the NaOH concentration determined from your titration curve: Holcag + NaOHeago - H₂ Ore) + NaCl caq? HCl + H₂O CI-t hot At equivalence pointe At end point& VB = 12.5mL VB = 13.4mL [H20+] = 10-pl [H₃O+] - 10-PH Chot] = 10-7.60 [H₂ot] = 2.3 4x10-12 mol [H₃O+] =2.51x10 - 8 mol 4-R1

Answer 1

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