CHM 116 Lecture 19: Addition Aspects of Equilibrium Continued 2
Document Summary
Naoh + hcl --> h2o + nacl (aq) Ph = 14 - 2. 04 = 11. 96. 0. 1 m of hc2h3o2 0. 1 m of naoh. Moles hcl = mhcl * vhcl = 0. 1 m * 0. 025 l = 00025 moles hcl. Moles naoh = mnaoh * vnaoh = 0. 1 m * 0. 005 l = 0. 0005 moles. The limiting reactant is naoh and they"re 1:1 ratio. Concentration: 0. 002 moles / 0. 030 l = 0. 06 m. Log(0. 06) = 1. 18 = ph after 5 ml of naoh. Moles naoh = mnaoh * vnaoh = 0. 1 m * 0. 010 l = 0. 0010 moles. There are 0. 0015 moles hcl remaining (naoh is still limiting) Concentration: 0. 0015 moles / 0. 035 l = 0. 0427 m. Moles naoh = mnaoh * vnaoh = 0. 1 m * 0. 020 l = 0. 002 moles naoh. There are 0. 0005 moles hcl remaining (naoh is still limiting) Concentration: 0. 0005 moles / 0. 045 l = 0. 01 m ph = -log(0. 011) = 1. 96.
