Question

4. Calculate [H30*], [OH-], pH and pOH of a 0.5 M CH3COOH solution, Ka = 1.8 x 10-5.

Answer 1

Glution CH3COOH solution is Chls COOH + H2O = CH. C00- + Hoot Equie boium constant (La) = (Chs Coo] [4304] - [ch cool ] In equilibrium constant Xa Water has been neglected because It is in huge amount do not play any role in equilibrium. any so Suka = [cH3.600] [Hz ot] [CH3coot] CH3COOH + H2O Hot + CH3600 0.5 Initial change fque brium - tu tn n 0.5-2 a Ka = nin 0.5-2 ka=1.8 X10

1.8x10-5 = x² (0.5-2) as compared too.5 en is very small value co we neglect the n from denominator So 0.5-2 written as 0.5. 3.8 X00-5 =_r? 0.5M B = (.8 410-5) (0.5) M n = √ 1.81105 x 0,5 M a = √ 1.8x5X106 m = 10 3/168X5 M = 10 39 M In = 3x103 30 concentration of (H3 0+) = 3610-3M . [H304 ) = 3 X1073 M Anso

We know that - Shot] [oH] = 1810-14 [04] = 3x10-14 [43 ot ] [OH = H10th 3x103 = 3.33 X10-12M Tot = 3.33X10 M ARS we know that POH = -log, [of ] and - P = loa va o*) (104 also PH+POH = 14 so PH = -log [ 30 t] (101 logio [3x103) = -log 10 6 -10910 61003) =-0.477 +3 --0.47773

Date Page NE PH = 2.523 pH = 2.523 ans. PH+POH = 14 POH = 14 - PH = 14- 20523 =11,477 POH = 11.477 Ang