Question

Impl e s Il Reeded lor this question. Determine the pH during the titration of 23.0 mL of 0.376 M hydrochloric acid by 0.310 M sodium hydroxide at the following points: (1) Before the addition of any sodium hydroxide (2) After the addition of 14.0 mL of sodium hydroxide (3) At the equivalence point (4) After adding 34.0 mL of sodium hydroxide Submit Answer

Answer 1

Hol vs Naoy Amount of Hel = 23 mLX 0.376 = 8.648 mmol o Amat of Naou = ompol. .: Insoln only, tee is proves [ut] = 0.376" I pu=0.42 |

37 ② After adding 14 mL naon : Amort of hel = 8.648 mmol Annt of Naoy = 1440.310 mool = 4.34 mnol Excen He= 4.00 mmol [ut] = 4 = 0.108M [pu = 0.966 At equiredence point Amont of nce- Anout of Neoon Complete neutraltion 3 mon 0 [Pu=7 | Only Nace is present - No hycholys's

② After adding 34 mL Noon Amout of her = 8.64 mnol Amrt of Noon = 10.54 mol Exen Naon = 1.9 mmol [ou] = 0.033 M poy = 1.48 IPY = 14-1.48 = 12.52