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scms/mod/ibis/view.php?id-5483829 rax Lab. Mercer Univerity-c.. x C What sThe Va

ID: 1046937 • Letter: S

Question

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 pA

Explanation / 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