Question

When a 17.3 mL sample of a 0.447 M aqueous hydrocyanic acid solution is titrated with a 0.377 M aqueous sodium hydroxide solution, what is the pH at the midpoint in the titration? pH = Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

"Immold HCN = comentation xvolume (in ml) - 17.3m1x0.447 m mmol of Naon = 0.377 MXV af midpoint of titration i.e. half equivalence point. here TH ON ] = ICN) »pH = pka t loq(CNI) Then = 9.240 pH=9.2 3 mmold HF = 36.4 mix 0.3911 = 14.23 mmol mmol de Naon = 0-25M X 17.9ml = 4.475mmol HF + H₂O - F + H₂of B 14.23 e -4.475 +4.475 +443 A 9.755 4.475 4.475 . THE]=9.755 - 9.755 M if- 4.475 38.4+17.9 543. 343 pu=pkat log (F-1 (MF) 53.2+109(4.975) pH = 2.86