A titration involves adding a reactant of known quantity to a solution of an ano
ID: 1012226 • Letter: A
Question
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) rightarrow A^- (aq) + H_2O(l) A certain weak acid, HA, with a K_a value of 5.61 Times 10^-6, is titrated with NaOH. A solution is made by titrating 9.00 m mol (millimoles) of HA and 3.00 m mol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 70.0 mL ? Express the pH numerically to two decimal places.Explanation / Answer
Part A:
One mole HA react with one mole NaOH gives one mole A- .
The mixture of weak acid HA and salt of its conjugage base (NaA) is a buffer.
pH = pKa + log[NaA]/[HA]}
pH = -log(Ka)+ log{(mmol NaA)/(mmol HA)}
pH = -log(5.61x10-6 )+ log{(3.00)/(6.00)}
pH = 4.95
Part B:
At equivalence point, mmol HA = mmol NaOH
Solution contains only conjugate base A-
[A- ] = (mmol NaA)/(total volume) = (9.00 mmol)/(70.0 mL) = 0.129 M
The reaction is,
A- + H2O <--------> HA + OH-
Suppose the xM A- is reacted.
Kb = {[HA][OH-]}/{[A- ]}
Kw/Ka = Kb
(1x10-14 )/(5.61x10-6 ) = x2/(0.129-x)
1.78x10-9 = x2 /0.129
x = 1.52x10-5 M
[HA] = 1.52x10-5 M and [A- ] = 0.129 M
pH = -log(5.61x10-6 )+ log{(0.129)/(1.52x10-5 )}
pH = 9.18
HA + NaOH ---------> NaA + H2O Initial(mmol): 9.00 3.00 - - Change(mmol): -3.00 -3.00 +3.00 Final (mmol): 6.00 0 3.00Related Questions
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