Question

A 38.9 mL sample of a 0.486 M aqueous nitrous acid solution is titrated with a 0.400 M aqueous barium hydroxide solution. What is the pH after 7.85 mL of base have been added?

pH =

Answer 1

of 0.486 HNOZ 38.9 mL Boco we have to find moles of acid present 0.400 M of Barium bydroside initially and moles of BoloH)2 moles Molarity volume in added. modes - molarity & volume in 2 3 0.486 mol x 38.9x10. moles of HNO2 L -0.01891 mole 3 L mole of Bach), added o.hou mol x 7:85x10 -0.00314 Dol Now, the balance equation for the reaction is 2 HNOG + Ba(OH)₂ - Ba NO2) + 2H₂O

BaOH)2 2 moles For I mole of @ of HNOL is required. remaining moles of HNO2 after reaction 0.01891 - 2X0-00314 = 0.01263 moles of Ba(NO2)2 formed = 0.00314 mol

as Now, BONO2)2 dissolves Ba(NO2) Be 2+ + 2 No2 So moes of No formed - 2x0.00314 -> 2 0.00628 mol Ooo pH of a So the solution after reaction contains 0.0126.3 mol UNO and 0.00628 mol NO2 These two care conjugate acid and base and makes the solution a buffer solation. buffer solution is given by Henderssen - Hasselbalch equation PH=pkat log kais dissociation Constant of acid is base here NO2 (АР is acid From tables, ka of HNOIS 7.1x104 (Please check the value of ka in your text book. Different textbook has different values like 4x104 5.6x104 pka = -logka = -log 7.1x154 _ 3.15 here HNO2 etc.) 7

3.15 & lo pH-pka tlog loge (6.00628 0.01263 CHA) pH - 2.85 (Please check the value of ka for HNO2 in the table in your your teactbook