Question

The pka of a weak acid is useful for identification of an unknown acid because it is a constant (as long as temperature remains the same). When a weak acid is titrated with a strong base, the pka of the weak acid equals the pH at the half-equivalence point. A 0.3210 g sample of weak monoprotic acid is delivered into laboratory glassware. It will be used for titratation against 0.1254 M NaOH. If the molar mass of the weak acid is 156.43 g/mol. What is the predicted equivalence point? (Do not enter units in your answers since they are already provided.) mL Complete the procedure for determining the pka of the weak acid. 1. Add deionized water to dissolve the 0.3210 g of weak acid 2. Add mL of 0.1254M NaOH to the acid 3. Mix thoroughly 4. Add indicator dye? (See next question) 5. Use a calibrated pH electrode to measure pH The resulting pH will equal the pka.

Answer 1

Moies of acid = 0.38 10g 156.43 Ilmol = 0.002052 moles CAH) Monoprotic HA + Nooh NaA + Hoces (Balanced) for I mole of ha, I mole of Nooh is required. for 0.00205 mole of Acid = 0.002054 mole of NowH required. Moles of Nowh= (Molarity) & (& NaoH) in Litre 0:00 205 mol = 0.1254 MX Nach Yeoh = 0.016354 Litre 0.016364 X1000 mL 16.364 ML Waon = * pka, Half neutralization is To Calculate required. For half - neutralization V Nash required = 16.364 mL = 8.182 m2