Question

Can anyone explain where they got 1.8 x 10^-5? How can that equal 1 if 0.5 and 0.5 cancel out?

EXl: What pH buffer of 0.50 M acetic acid and of a 0.50 M sodium acetate? is the comprised 1x107 M in neutral water 232 (aq) 0.50 +X I 0.50 C-X E 0.50-X K,= ~0 0.50 + X [HC2HO2-X] [0.50-x [X][0.50] 区110,501 ~ 1.8x10 [0.50] Xe[H+]= 1.8 x 10-5 M pH = 4.74 X is <<5% of 0.50 (it is 0.0036%), so simplifying assumption is valid

Answer 1

Ka of acetic acid=1.8x10^-5

As per the given ICE chart

Ka=$\frac{[H_3O^&plus;][C_2H_5COO^-]}{[CH_3COOH]}$=$\frac{x\times (0.50&plus;x)}{(0.50-x)}$=1.8x10^-5

As Ka is very small so the dissociation of the acetic acid is very small and hence value of x is very small as compared to 0.50 M and can be ignored with respect to 0.50 M.

Ka$\approx$$\frac{x\times (0.50)}{(0.50)}$=1.8x10^-5

x$\approx$1.8x10^-5=[H₃O⁺]

pH=-log[H₃O⁺]=-log 1.8x10^-5=-0.26+5=4.74