What is the % pH change if a student added 0.0025 mol of Ca(OH)2 to a 1.00 L buffer solution containing 0.100 M HC2H3O2 and 0.200 M NaC2H3O2.
pH = pKa - lg"\\frac{C(acid)}{C(salt)}"
pKa(HC2H3O2 ) = 4.75
pH1 = 4.75 - "lg\\frac{0.1}{0.2}" = 5.05
C2(acid) = 0.1 - 0.0025 "\\times" 2 = 0.095 M
C2(salt) = 0.2 + 0.0025 "\\times" 2 = 0.205 M
pH2 = 4.75 - lg"\\frac{0.095}{0.205}" = 5.86
"\\Delta"pH = 5.86 - 5.05 = 0.81
%pHchange = "\\frac{0.81 \\times 100}{5.05}" = 16 %
Answer: 16 %.
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