A52.0-ml volume of 0.35 m ch3cooh (ka=1.8×10−5) is titrated with 0.40 m naoh. calculate the ph after the addition of 17.0 ml of naoh. express your answer numerically.

2.76.

Explanation:

C of acid = [(MV)CH₃COOH - (MV)NaOH] / Vtotal.

C of acid = [(0.35 M)(52.0 mL) - (0.4 M)(17.0 mL)] / (69.0 mL) = 0.165 M.

∵ [H⁺] = √(Ka.C)

∴ [H⁺] = √(1.8 x 10⁻⁵)(0.165 M) = 1.72 x 10⁻³.

∵ pH = - log[H⁺] = - log (1.72 x 10⁻³) = 2.76.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.