Chemistry - calculating concentrations (Molarity) equilibrium. Iteration or Quad

Question

Chemistry - calculating concentrations (Molarity) equilibrium. Iteration or Quadratic formula? Getting different answers?

Let x = [H2SO3] that ionizes to form [H3O+] = [HSO3-]
       Ka1 = 1.7 x 10-2 = ([H3O+][HSO3-]) / [H2SO3]
           1.7 x 10-2 = (x)(x) / 0.100-x

Assuming the x is small as compared to 0.10, find x = 0.0412M
      but, when you check this our, this is 41.2% of 0.100M; therefore, have to find a better value of x.

ITERATE:
            1.7 x 10-2 = (x)(x)/0.100 - 0.0412
       Recalculate  x = 0.0316M. Iterating one more time,
            1.7 x 10-2 = (x)(x)/0.100 - 0.0316
        Now, x = 0.0341M.    Iterating one more time,
             1.7 x 10-2 = (x)(x)/0.100 - 0.0341
        Now, x = 0.0336 M =  [H3O+] = [HSO3-]
   If we iterate one more time, we will again get 0.0336M

When I use the quadratic formula:

(1.7x10^-2) + ?(.000289-4 (-.0017) )

/ 2

I get .0505981, not .0336. What is wrong here?

Explanation / Answer

When using quadratic formula, you get two values, typically a positive and a negative one. The one you showed me, .05059 is the NEGATIVE value. Negative concentrations DOES NOT exist, therefore you need to take it out of the solutions. The other positive solution is, in fact, 0.0335 this is your solution!

Let me solve you the quadratic formula:

ax^2 + bx + c = 0

Let

a= 1

b = 1.7*10^-2

c= -(0.1)*(1.7*10^-2)

x = [-b+/-sqrt(b^2-4ac) ]/2a

x = [-(1.7*10^-2)+/- sqrt[(1.7*10^-2)^2-4*1*(-0.1)(1.7*10^-2))]/(2*1)

x= -0.0505 and +0.0336

Once again, choose the positive solution since there are no negative concentrations

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.