Question #46e01

1 Answer
Al E.
Jan 2, 2018

Consider the neutralization,

#CH_3COOH(aq) + OH^(-)(aq) to CH_3COO^(-)(aq) + H_2O(l)#

puu.sh

Now, consider the equilibrium at the end point of acetate, the conjugate base of acetic acid,

#CH_3COO^(-)(aq) +H_2O(l) rightleftharpoons OH^(-)(aq) + CH_3COOH(aq)#

puu.sh

where, #K_b = ([OH^-][CH-3COOH])/([CH_3COO^(-)]) = 5.6*10^-10#

Hence,

#5.6*10^-10 = x^2/(0.0967)=> x =[OH^-] approx 7.33*10^-6M#
#therefore "pH" = 14+log[OH^-] approx 8.87#

This is reasonable, because only the conjugate base is present at the end point. Giving it room to increase the pH of the solution to a slightly basic level.

Note: the BCA table is using moles, and the ICE table is using molarity (assuming that volumes are additive).

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