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Answer #1

MW of NaC7H5 O2 -144.11g/mol wt = 10.06 g

mol of NaC7H5O2 = 10.06 /144.11=0.0698mol

con of sodium benzoate= 0.0698 / 2.5 = 0.0279M

benzoic acid = 0.360M

ka= 6.3x10^-5 pKa = -log Ka = -log 6.3 x10^-5 = 5- 0.79934=4.20066

pH = pKa + log[salt] / [acid] = 4.20066 +log [0.0279]/ [0.36]=4.20066+log 0.0775=4.20066- 1.11069

pH= 3.08997

when HCl is added

HC7H5O2 + H+   --------> C7H5O2-

0.9 mol         0               0.0698 mol initial

0.9 + 0.0125mol           0.0698 - 0.0125 at equlibirium

pH = pKa + log [salt] / [acid ]

pH = 4.20066+ log [0.0573]/ [0.9125]

pH = 4.20066+log 0.06279452

pH = 4.20066 -1.20207= 2.99859

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