Question

late the pH of a mixture that is 0.095 M in HF and 0.225 M in HC6H5O.

Answer 1

mixture of 0.095M HF and O.da5M (64504 dissociate Both HF and Pa and na Consoh 6hSon are wea are weak acid ka of 4F = 7.2810" Raj C6450H = 1.6810-10 kag HF >> Kaf long oth so in the presence of HF the dissociation of C645o4 furthur depressed due to common in effect.

| 2 HF | 0.075 H+ +F- o o н x € 0.015-x о tx x to x ka= [ht] [F] [IF ] (x) (x) (0.015-2) 7.2410-4 = x <co.015

7-24104 ? 0.095 6.84x10% = x² 6.843785 X 0.0083 M = x X = 0.0083 [F] = [4+] = 0.0083 gholt (64500 Now consider dissociatin C6450 447 (64504 4 0.225 ICM) tx tx 0.00f3+x 0.225-* Ka= [he] [Conso-] (сочу?"] 3) (0.0083+) 70.225-) = x<c0.125 1.6X10t0 = (x) 0.0083 0.225 4.34316" - 5 go (11) = 1 + 0.0003 = 4:34X10-9 +0.0081: 0.0083 ph= -log[ut] E-boy [0.2013) pu= d.os pu of solution = 2.08 Answer