Question

| | a search signmen 110% Review Topics References Use the References to access important values if needed for this question. A 48.7 mL sample of a 0.540 M aqueous hypochlorous acid solution is titrated with a 0.397 M aqueous sodium hydroxide solution. What is the pH after 45.4 mL of base have been added? pH- Submit Answer Rety Entire Group 1 more group attempttemaining

Answer 1

number of moles of HCl = 48.7*0.54/1000 = 0.0263 moles

number of moles of NaOH = 45.4*0.397/1000 = 0.01802 moles

NaOH + HCl -----> NaOH + H2O

1 mole of NaOH requres 1 mole of HCl

0.01802 moles of NaOH requires 0.01802 moles of HCl

number of moles of HCl remaining = 0.0263 - 0.01802 = 0.00828

concentration of H+ ions = 0.00828*1000/(48.7+45.4) = 0.088

pH = -log[H+]

pH = -log(0.088)

pH = 1.056