Question

Tuo UF Weak Acid with Strong Base 5 of 7 > A certain weak acid, HA, with a Ka value of 5.61 x 10 Constants Periodic Table A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is HA(aq) + OH (aq) + (aq) + H2O(1) is titrated with NaOH. Part A A solution is made by titrating 8.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. What is the resulting pH?! Express the pH numerically to two decimal places. ► View Available Hint(s) O AO O 2 ? pH = Submit Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point it the total volume is 65.0 mL ?

Answer 1

HA + CH F 04 Á + H₂O (42) 199) (aq) 8 milliard 3 millinel - - +3 milling - millinel - 3 millime H +5 millime +3 milline . HA Qq, & á lag, will form Buffer soutim whose pr is calculated by Henderson equation H ka P 102 " = -6 LHA ka =561410 -6 pika = -log(5.61x10)=5.25

pH = 5.25 + log ) = 5.25 + (-0.22) = 5:03