Question

Calculate the pH of a 0.500 L solution of 0.800 mol NaCH3CO2. Ka for acetic acid, CH3CO2H, is 1.8 x 10:5. 2.27 9.47 ooo 11.73 4.53

Answer 1

Ans CH₂ CO2 H + H₂O = CH₂ CO2 ans + H30h (as.) ca =) =) ka. ks = kw=1014 (1.8 x 105) (Kb) = 1014 kb - 10" Kb = TOIL -10 5.5 x 10 1.8X105 Here we have - Na CH₃CO2 - 0.5002 and 0.500 mol As we know- conc" = moles Blume (2) i conch of NaCH₃CO2 = 0.800 = 1.6m 0.500 Dissociation of NaCH3CO₂ (It is salt : completely dissolve) NaCH₃CO2 las.) (Na (as) + CH₃C00 cas) 1-6M 1.6M Doesn't account for pH Now, CH₂ coo will hydsolise in water as Kon CH3200 Hang + OH (ar) D (H3100 g) + H20 (1) 1.6 ộ – a tu 1.6-x

kb = } [CH₃00H] [ori [CH₂000) (x) (x) (1.6-2) 5.5 X 1070 x2 I since value of Ky is very l small, thus we can take- 1.6-x ñ 1.6 1.6 =) 2 = 2.97 xos Thus, cal equilibrium. TOH) = x=2.97x105 m and Ipon=- log [Ort] - log (2.97x108)=4.53 also, TpH = 14-poh 14-4:53= 9.47 Hence, option (6) 9.47 is CORRECT Please Rate ...