A sample of 0.2505 g of an unknown monoprotic acid...

Question

A sample of 0.2505 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M . The acid required 30.0 mL of base to reach the equivalence point.

After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the K_a for the unknown acid?

Jules Bruno · Accepted Answer

We’re being asked to calculate the Ka of the given monoprotic acid. Recall that at the equivalence point of a titration:moles acid=moles baseAlso, recall that moles = molarity × volume. This means:MVacid=MVbaseThe monoprotic acid will have 1 equivalence point:HA → H+ + A-                                  (MA)acid = (MV)baseWe are given:[HA]=0.0950 MVolume of HA = 25 mLVolume of A- to reach equivalence point=30 mLRecall: A-   1 mL = 10-3 L[readmore]mol acid = mol base at equivalence point =30 mL(10-3 L1 mL)(0.0950 molL)At equivalence point, mol acid = mol base = 2.85×10-3 molRecall:Ka=[products][reactants]=[H+][A-][HA]H+ is used interchangeably with H3O+To determine [H+], we can derive it from the pH using the formula for the determination of pH.pH=-log[H+]To get rid of the log on the one side we can raise both sides of the equation to 10.[H+]=10-pHWe have a pH of 6.50, hence we have an [H+] of:[H+]=10-6.50=3.16×10-7 MFor [HA] we can obtain this by subtracting the moles of the base from the moles of acid then divide by total volume at equilibrium.We are given:Total Volume=Volume of Base + Volume of Acid = 15 mL + 25 mL = 40 mL[HA] = 0.0950 Mmol HA = 2.85×10-3 mol[HA]=(2.85×10-3 mol)-(15 mL×10-3 L1 mL)(0.0950 molL)40 mL(10-3 L1 mL)=0.035625 MFor [A-] we can obtain this by determining the moles of the base and divided by total volume at equilibrium.[A-]=(15 mL)(0.0950 molL)40 mL=0.035625 MSubstituting these value to the Ka expression yields the value for Ka:Ka=[products][reactants]=[H+][A-][HA]Ka=[H+][A-][HA]=(3.16×10-7 M)(0.035625 M)(0.035625 M)=3.16×10-7 MAlternatively, at the half-equivalence point, there is simply one-half of the equivalence point. The volume needed to reach the half-equivalence point is simply half of the volume required to reach the equivalence point (i.e. 15 mL is the half-equivalence point since the volume necessary to reach equivalence point is 30 mL).At ½ the equivalence point:pH=pKaWe have a pH of 6.50pKa= 6.50To get the Ka value, we can use the formula below:Ka=10-pKaKa=10-6.50Ka = =3.16×10-7 M

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