Quadratic Solutions (continued) For the equilibrium HF (aq) - H (aq) F (aq) (c)
ID: 704908 • Letter: Q
Question
Quadratic Solutions (continued) For the equilibrium HF (aq) - H (aq) F (aq) (c) If the Kc at room temperature is 6.80 x 104, calculate H] in a 0.0300 M HF solution using the quadratic solution: (d) Calculate the [H] in a 0.300 M solution without using the quadratic solution (e) Calculate the H'] in a 3.00 M solution using the quadratic solution (f) Complete the following table, where [HFlo means original or starting concentration: [HFo Non-Quadratic Non-Quadratc Quadratic HTError [H1 Amount(Correct) [H' Quadratic vs. ionized (06) Non-Quadratic 0.0300 0.300 3.00 4.52 x 103 1.39 x 102 4.52 x 10Explanation / Answer
for the reaction HF<-------->H++F-
KC= [H+] [F-]/[HF]
Preparing the ICE Table
initial concentration change in concentration Equilibrum concentration
HF M -x M-x
H+ 0 x x
F- 0 x x
Kc= x2/(M-x)= 6.8*10-4 (1)
x2+6.8*10-4x - M*6.8*10-4 =0
This is a quadratic equation of the form ax2+bx+C=0, whose roots are x= -b(+-)sqrt(b2-4ac)/2a
only one root is valid and gence x= -b+sqrt(b2-4ac)/2a
b=6.8*10-4, a=1 and C= -M*6.8*10-4, x={-6.8*10-4+ sqrt(6.8*10-4*6.8*10-4+4*1*M*6.8*10-4)}/2 (2)
when M=0.03M, Eq.2 becomes x= [H+]=0.0042, pH= -log [H+], pH=2.4
when M=0.3M, Eq.2 becomes x=0.0139, pH=1.855
when M=3M, Eq.2 becomes, x=0.0448 and PH= 1.35
when M is >>>x, Eq.1 becomes x2/M= 6.8*10-4, x= sqrt(6.8*10-4M) (3)
when x= 0.03M, Eq.3 becomes x= sqrt(6.8*10-4*0.03)= 0.0045, % error quadratic vs non- quadratic = 100*{(0.0042-0.0045)/0.0042}=-7.14%, percent ionized = 100*0.0045/0.03= 15%
when x=0.3M, Eq.3 becomes, x=0.0143, % error= 100*{(0.0139-0.0143)/0.0139}=-2.8%, percent ionized= 100*0.0143/0.3= 4.76%
when x=3M, Eq.3 becomes x=0.0452, % error= 100*{(0.0448-0.0452)/0.0448}=-0.89%, percent ionized= 100*0.0452/3= 1.51
when x=3M, E