scms/mod/ibis/view.php?id-5483829 rax Lab. Mercer Univerity-c.. x C What sThe Va
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scms/mod/ibis/view.php?id-5483829 rax Lab. Mercer Univerity-c.. x C What sThe Value Of KF.. Y Whenthe oude of genet Streaming Movies gener... . Co Ca Sapling Learning Mapd To a 25.00-mL volumetric flask, a lab technician adds a 0.150 g sample of a weak and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with M KOH. She reaches the endpoint after adding 43.41 mL of the KOH solution. acid, HA, 0 .0888 Determine the number of moles of the weak acid in the solution. Number 3.85x10 3 mol Determine the molar mass of the weak acid Number 1.96 mol R/mol After the technician adds 15.71 mL of the KOH solution, the phH of the mixture is 4.49 Determine the pK, of the weak acid pAExplanation / Answer
number of moles of KOH = 0.0888 * 43.41 = 3.855 mmoles
HA + KOH ------> KA + H2O
1 mole of KOH requires 1 moles of HA so,
number of moles H+ ion reacted in the titration = 3.855 * 10^-3 moles
number of moles of HA = 3.855 * 10^-3 moles
molar mass of acid = 0.15/(3.855*10^-3) = 38.91 g/mol
number of moles of KOH = 15.71 * 0.0888 = 1.395 mmoles
pH = pKa + log[salt]/[acid]
4.49 = pKa + log(1.395/(3.855-1.396))
pKa = 4.736
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