Question

please answer its parts

Determine the pH change when 0.111 mol NaOH is added to 1.00 L of a buffer solution that is 0.379 M in HF and 0.350 M in F. pH after addition - pH before addition = pH change =

Answer 1

3.18 i) Ka of Hf = 6.6810-4 pkaa Log (6.6810-4) [HA] = 0.379 M [F] : 0.85 M :. pH = bopka + log [eo] CHF] 3.18 + log ( 0.35 0.37 ) : 3.145 0.1ll mol NaOH is added Hft NaoH • Naf + H₂O 0.379 mol 0.111 mol 0.35 mol H -0.lll mot 0.111m of +0.111 mol E 0.268 mol o 0.46 mol Thew, [ue] 0.268 mol 0.268M IL [Nar] 0.461 mot 0.461M IL 10.461 pH = 3.18 + log 0.268 : 8.415 :. Again pH change 3.415 - 3.145 : 0.27 CS$canned with CamScanner

m) [CH, NH3 CL ] = 0.305 M [CH3NH₂] = 0.246 pka of CH ₂ NH g t = 4.17 :: PHP 4.17+ log 16-241 . a 4.076 0.305 CH3NH₂ U + NaOH > CH₃NH₂ + NaCl + H2O 0.063 md 0.305 mol 1 0.246 m od C -0.063 mol 0.063 mol +0.06 mol 0.309 mol o 0.242 mol الب 0.242ml Ther, [CH3NH24] 0.242M IL 0.309 mol 0.309M [CH₂ Nik] Լ 0.309 ::. pH= 4.12 + log (:. 1 ) :.4.2276 0.242 0.2 Ther, change pH = 4.276-4.076 CS$canned with CamScanner

3) [NaHSO₃] = 0.363 M [K2 503 ] 0.376 pka of NaHSO3 = 9.7 0.376 PH = -3.7 + log (a 3.684 0.363 Na H50₂ +KOH → K₂SO₂ Na + H20 1 0.363 mol 6.089 mol 0.376 mol с + 0.089 mol 0.089 mol 0.089 m of 0.465 mol 0 E 2 0.274 mol см, 0.2744 . [na H50,]: 0.274 mol 14 [ K₂ 50₂] 6.465 ml IL 0.465 M 0.465 : 3.47 :.PH 3.7 + log 0.274 Again 3.47 to 3.684 : 0.214 pH change CSScanned with CamScanner