Question

An analytical chemist is titrating 123.5 mL of a 0.9200 M solution of nitrous acid (HNO 2) with a 0.2100 M solution of NaOH. The p K, of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 230.0 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places.

Answer 1

lue know the neutralization mearton between Neak arad (thlo) and strong base (a 244) Halo thach Mano of the (salt) ere I mole of Naoff will neutralère I mole of Halo to form 1 onoles of salt Hallo 11 11 Maot Now geran- Esiti al concentrat en of Malotaken (Mo) 20.92001 satial (Mi) = 0.2100M Enitial volume of Healey taką (1); 123.5 ol initial NAOH How te foral volume after additum of two (123.5 + 230) onl 1H (V;) = 230 od - 1353.5 ond] So the new shitial concentration of species (M) after addition of beeth is - [ANO,J= Miri 353.5 onl [Ma Coty Mini 0.9200 MX 123.5 ord 10.3214 Vz oil @- 2100 x 230 353.5 onl 04137 m 2 V2

So the equilibrium concestonteren of the special will be Reaction → Nalla Halo, t Ma oll 0:137M 0.3241 O -0.137 M change Equile bring -0.137 M +0.137m 0 1841 0:1371 are Here we can see, at equiliinam, her hair both the weak arod conlee) and the salt of its conjugate base allo) present. So the solution is buffer. And the pH of abutter is calculated by Henderson Hasselbach equation (salt) pH a phat logo Canoel = 3.35 + log 0.87M 0.184m 23.25 t log (0.745). 0-13 -3.25 TpH 1) 3.22