Question

The base protonation constant Kb of lidocane (C14H21NONH) is 1.5x10^-8.

Calculate the pH of a 3.33 M solution of lidocane at 25 degrees Celcius. Round to 1 decimal place.

Answer 1

The question gives us the Kb and the initial concentration. So step 1 is to find the equilibrium concentrations of lidocane, its conjugate acid using an ICE table.

One the equilibrium concentrations have been evaluated, we can use the henderson hasselbach equation to calculate pOH and then pH.

I have shown all steps in the image attached below. Please do give a thumbs up is I helped. Thanks :).

Step 1:- Calculate the equilibrium concentrations of lidocane, it's conjugate and and onion. Lidocane exists is a weak base equilibriom with so it dissociates its conjugate acid parrally and and ch :- in CIAH, NONH TH₂O - CAH,, NONH Tor Construct an ICE Table [cia H₂i NONH] | [cuha,NONHA] | Cont] 1. 0:00 Initial conco 3.33 M change . X Equilibrium (3:33=) M conc: ill . ;. . () (2) K = L Cu Hai NONH, J Lour 3.33-6 1 [Cutz, Nont] x is small حد ما نلحيط small so we can neglett 1o 5 x 10 = 3.33 => x= 0.00022 M [cun.. NONH, J = 0.00022 LCGY HEINONH] = 3.33 -0.00022 L (GUH2. Henderson-Hasselbach equation to find the Step 2 - use the - pH. 21 NO We need: pk = -log ko = 7.824 cil) [cula, NONH] = 3.32976m Cli) [CluHz, NONH, ] = 0.00022M police Pol=pky + log ] [auth, NONH [Cutz, NONH = 7.8247.log 60.00022) - 3.644 . (3.32978) CSsganned with fansmanned => pH= 14-pon = 14-3-644 = 110.41 Also ph

Note: While calculating the equilibrium concentration I used an assumption. Since Kb has a very small value, x will also have a small value. So it can be neglected in the denominator as shown in the image.